5.11 Solving problems in mechanics

The three laws of motion that you have learnt in this chapter are the foundation of mechanics. A typical problem in mechanics usually does not merely involve a single body under the action of given forces. More often, we will need to consider an assembly of different bodies exerting forces on each other. Besides, each body in the assembly experiences the force of gravity. When trying to solve a problem of this type, it is useful to remember the fact that we can choose any part of the assembly and apply the laws of motion to that part provided we include all forces on the chosen part due to the remaining parts of the assembly. We may call the chosen part of the assembly as the system and the remaining part of the assembly (plus any other agencies of forces) as the environment. We have followed the same method in the solved examples. To handle a typical problem in mechanics systematically, one should use the following steps:

(i) Draw a diagram showing schematically the various parts of the assembly of bodies, the links, supports, etc.

(ii) Choose a convenient part of the assembly as one system.

(iii) Draw a separate diagram which shows this system and all the forces on the system by the remaining part of the assembly. Include also the forces on the system by other agencies. Do not include the forces on the environment by the system. A diagram of this type is known as ‘a free-body diagram’. (Note this does not imply that the system under consideration is without a net force).

(iv) In a free-body diagram, include information about forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion.

(v) If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of \(A\), the force on \(A\) due to \(B\) is shown as \(\mathbf{F}\), then in the free-body diagram of \(B\), the force on \(B\) due to \(A\) should be shown as \(-\mathbf{F}\).

Example 5.19: A wooden block of mass \(2 \mathrm{~kg}\) rests on a soft horizontal floor. When an iron cylinder of mass \(25 \mathrm{~kg}\) is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of \(0.1 \mathrm{~m} \mathrm{~s}^{-2}\). What is the action of the block on the floor (a) before and (b) after the floor yields? Take \(g=10 \mathrm{~m} \mathrm{~s}\). Identify the action-reaction pairs in the problem.

Solution: 

(a) The block is at rest on the floor. Its free-body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to \(2 \times 10=20 \mathrm{~N}\); and the normal force \(R\) of the floor on the block. By the First Law,the net force on the block must be zero i.e., \(R=20 \mathrm{~N}\). Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to \(20 \mathrm{~N}\) and directed vertically downwards.
(b) The system (block + cylinder) accelerates downwards with \(0.1 \mathrm{~m} \mathrm{~s}^{-2}\). The free-body diagram of the system shows two forces on the system: the force of gravity due to the earth \((270 \mathrm{~N})\); and the normal force \(R^{\prime}\) by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system,
\(
270-R^{\prime}=27 \times 0.1 \mathrm{~N}
\)
ie. \(R^{\prime}=267.3 \mathrm{~N}\)
By the third law, the action of the system on the floor is equal to \(267.3 \mathrm{~N}\) vertically downward.
Action-reaction pairs
For (a): (i) the force of gravity ( \(20 \mathrm{~N})\) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to \(20 \mathrm{~N}\) directed upwards (not shown in the figure).
(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).

For (b): (i) the force of gravity \((270 \mathrm{~N})\) on the system by the earth (say, action); the force of gravity on the earth by the system (reaction), equal to \(270 \mathrm{~N}\), directed upwards (not shown in the figure).
(ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

Example 5.20:A box of \(5.0 \mathrm{~kg}\) stays on a frictionless horizontal surface.
Sham then starts pulling the box with a force of \(100 \mathrm{~N}\) at an angle of \(15^{\circ}\) with the horizontal.
a. Draw a free-body diagram for the box.
b. What is the resultant force acting on the box?
c. What is the acceleration of the box?
d. What is the normal force that the surface exerts on the box?

Solution: Before solving this problem, we probably need to draw a horizontal surface, a rectangular box sitting on the surface, and indicate that the box is being pulled by a force that makes \(15^{\circ}\) with the horizontal:

The next step is to look carefully at our sketch and try to enumerate all the forces that are acting on the box.
There are clearly three forces acting on the box:

• Sham’s pulling force, which we’ll indicate with \(\mathbf{F}\)
• The gravitational force \(m \mathbf{g}\)
• and the normal force \(\mathbf{N}\), which the surface exerts to prevent the penetration of the box
  At this point we are ready to draw a free-body diagram for the box:

We know the mass of the box \((5.0 \mathrm{~kg})\) and Bob’s pull \(\left(100 \mathrm{~N}\right.\) at \(15^{\circ}\) angle with the horizontal).
The problem is asking us to draw a free-body diagram for the box (which we already did), then to determine the resultant force that acts on the box, the acceleration that the box has as a result, and the normal force that the surface exerts on the box.
In short:
We know
\(
\begin{aligned}
&m=5.0 \mathrm{~kg} \\
&F=100 \mathrm{~N} \\
&\theta=15^{\circ}
\end{aligned}
\)
We want to know
\(
\begin{aligned}
&\mathbf{R}=? \\
&\mathbf{a}=? \\
&\mathbf{N}=?
\end{aligned}
\)

We can start by finding the resultant force. So, let’s draw the coordinate axes on our diagram, with the \(x\) axis in the direction of motion, and find the \(x\) and \(y\) components of the 3 forces that act on the box:

\(
\begin{array}{ll}
F_x=F \cos 15^{\circ} & F_y=F \sin 15^{\circ} \\
N_x=0 & N_y=N \\
m g_x=0 & m g_y=-m g
\end{array}
\)
We can now find the \(x\) and \(y\) components of the resultant force:
\(
\begin{aligned}
&R_x=F_x+N_x+m g_x \\
&R_x=F \cos 15^{\circ}+0+0 \\
&R_x=F \cos 15^{\circ} \dots(1)
\end{aligned}
\)

\(
\begin{aligned}
&R_y=F_y+N_y+m g_y \\
&R_y=F \sin 15^{\circ}+N+(-m g) \\
&R_y=F \sin 15^{\circ}+N-m g \dots(2)
\end{aligned}
\)

Since the motion is along a horizontal surface, the \(y\) component of the resultant force, \(R_y\) must be zero. Otherwise, there would be an acceleration in the \(y\) direction (for Newton’s \(2^{\text {nd }}\) Law), which is not the case.
\(
R_y=0
\)
Therefore, the magnitude of \(\mathbf{R}\) will be equal to the magnitude of its \(x\) component.
The \(x\) component, \(R_x\), is positive because the box is accelerating in the positive \(x\) direction:
\(
\begin{aligned}
&R_x>0 \\
&R=R_x
\end{aligned}
\)
We have already determined \(R_x\) in Eq. (1):
\(
\begin{aligned}
&R_x=F \cos 15^{\circ} \\
&R=R_x=F \cos 15^{\circ} \\
&R=(100 \mathrm{~N})\left(\cos 15^{\circ}\right) \\
&R=97 \mathrm{~N}
\end{aligned}
\)
And \(\mathbf{R}\) will have direction in the positive \(x\) axis:

Next, let’s find the acceleration of the box.
We know the mass of the box, and we just found the resultant force acting on the box. We can apply Newton’s \(2^{\text {nd }}\) Law to get the acceleration:
\(
\begin{aligned}
&\mathbf{R}=m \mathbf{a} \\
&\mathbf{a}=\frac{\mathbf{R}}{m}
\end{aligned}
\)
The direction of \(\mathbf{a}\) is the same as that of \(\mathbf{R}\). Let’s calculate the magnitude of \(\mathbf{a}\) :
\(
\begin{aligned}
&a=\frac{R}{m} \\
&a=\frac{97 \mathrm{~N}}{5.0 \mathrm{~kg}} \\
&a=19 \mathrm{~N} / \mathrm{kg}=19 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
The last unknown on our list is the normal force \(\mathbf{N}\).
\(\mathbf{N}\) has direction in the positive \(y\) axis. But what about the magnitude?
We saw \(N\) appear in Eq. (2):
\(
R_y=F \sin 15^{\circ}+N-m g
\)
And we know that \(R_y\) is zero, therefore:
\(
0=F \sin 15^{\circ}+N-m g
\)
This is an equation with one unknown \((N)\). So, let’s solve it:
\(
\begin{aligned}
&0=F \sin 15^{\circ}+N-m g \\
&-N=F \sin 15^{\circ}-m g \\
&N=-F \sin 15^{\circ}+m g \\
&N=m g-F \sin 15^{\circ} \\
&N=(5.0 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})-(100 \mathrm{~N})\left(\sin 15^{\circ}\right) \\
&N=49 \mathrm{~N}-26 \mathrm{~N} \\
&N=23 \mathrm{~N}
\end{aligned}
\)
We have found all the unknowns the problem asked us:
\(
\begin{aligned}
&R=97 \mathrm{~N} \\
&a=19 \mathrm{~m} / \mathrm{s}^2 \\
&N=23 \mathrm{~N}
\end{aligned}
\)

Example 5.21: Frank is pushing a block of \(50 \mathrm{~kg}\) over the floor, with a force of \(600 \mathrm{~N}\) downward and forward, making a \(20^{\circ}\) angle with the horizontal. The coefficient of sliding friction between the block and the floor is \(0.39\).
Find the acceleration of the block and the friction force acting on the block.

Solution:

We need to draw a horizontal surface, a block on that surface and indicate that the block is pushed by a downward-forward force that makes \(20^{\circ}\) with the horizontal. We also need to indicate that there is friction between the block and the floor:

Next, we look at our sketch and try to list all the forces that act on the block.
There are four forces acting:

• Frank’s pushing force, which we’ll indicate with \(\mathbf{F}\)
• the friction force, \(\mathbf{F}_{\mathrm{f}}\)
• the force of gravity, \(m \mathbf{g}\)
• and the normal force \(\mathbf{N}\), which the surface exerts on the block
  Knowing the forces, let’s draw a free-body diagram for the block:

Let’s take a step back and examine what we know, and what the need to find:
We know the mass of the block (50 kg), the push ( \(600 \mathrm{~N}\) at \(20^{\circ}\) downward-forward), and the coefficient of friction (0.39).
We need to find the acceleration of the block and the force of friction.
We know
\(
\begin{aligned}
&m=50 \mathrm{~kg} \\
&F=600 \mathrm{~N} \\
&\theta=20^{\circ} \\
&\mu=0.39
\end{aligned}
\)
We want to know
\(
\begin{aligned}
&\mathbf{a}=? \\
&\mathbf{F}_{\mathrm{f}}=?
\end{aligned}
\)
Since we know the mass of the block, if we first find the resultant force, we can then apply Newton’s \(2^{\text {nd }}\) Law to get the block’s acceleration.

Let’s start by drawing the coordinate axes on our free-body diagram, and determine the \(x\) and \(y\) components of all the forces acting on the block.

\(
\begin{array}{ll}
F_x=F \cos 20^{\circ} & F_y=-F \sin 20^{\circ} \\
F_{\mathrm{f}_x}=-F_{\mathrm{f}} & F_{\mathrm{f}_y}=0 \\
N_x=0 & N_y=N \\
m g_x=0 & m g_y=-m g
\end{array}
\)
We can now find the \(x\) and \(y\) components of the resultant force:

\(
\begin{aligned}
&R_x=F_x+F_{\mathrm{f}_x}+N_x+m g_x \\
&R_x=F \cos 20^{\circ}+\left(-F_{\mathrm{f}}\right)+0+0 \\
&R_x=F \cos 20^{\circ}-F_{\mathrm{f}} \dots(1)
\end{aligned}
\)

\(
\begin{aligned}
&R_y=F_y+F_{\mathrm{f}_y}+N_y+m g_y \\
&R_y=\left(-F \sin 20^{\circ}\right)+0+N+(-m g) \\
&R_y=N-F \sin 20^{\circ}-m g \dots(2)
\end{aligned}
\)
For Eq. (1) we need to find the friction force \(F_{\mathrm{f}}\).

By definition, the direction of the sliding friction force will be opposite to the motion of the block, and its magnitude will be equal to the product between the coefficient of sliding friction and the normal force:
\(
F_{\mathrm{f}}=\mu N
\)
We know \(\mu\), but we don’t know \(N\).
Where did we see \(N\) before?
We saw \(N\) in Eq. (2):
\(
R_y=N-F \sin 20^{\circ}-m g
\)
Since the block is moving along the floor, \(R_y\) must be zero (otherwise there would be a vertical acceleration):
\(
R_y=0
\)
Therefore:
\(
0=N-F \sin 20^{\circ}-m g
\)
The only unknown in this equation is \(N\), so we can solve it:
\(
\begin{aligned}
&0=N-F \sin 20^{\circ}-m g \\
&N-F \sin 20^{\circ}-m g=0 \\
&N=F \sin 20^{\circ}+m g
\end{aligned}
\)
Having found \(N\), we can calculate \(F_f\) :
\(
F_{\mathrm{f}}=\mu N
\)
\(
\begin{aligned}
&F_{\mathrm{f}}=\mu\left(F \sin 20^{\circ}+m g\right) \\
&F_{\mathrm{f}}=(0.39)\left[(600 \mathrm{~N})\left(\sin 20^{\circ}\right)+(50 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})\right] \\
&F_{\mathrm{f}}=(0.39)(205 \mathrm{~N}+490 \mathrm{~N}) \\
&F_{\mathrm{f}}=270 \mathrm{~N}
\end{aligned}
\)
Now that we know \(F_{\mathrm{f}}\), we can find \(R_x\) from Eq. (1):
\(
R_x=F \cos 20^{\circ}-F_{\mathrm{f}}
\)
\(R_x\) is positive (since the block accelerates in the positive \(x\) direction), and we have shown that \(R_y\) is zero. This means that \(\mathbf{R}\) will have direction in the positive \(x\) axis, and its magnitude will be equal to \(R_x\) :
\(
\begin{aligned}
&R=R_x \\
&R=F \cos 20^{\circ}-F_{\mathrm{f}}
\end{aligned}
\)
Now that we finally found the resultant force, let’s apply Newton’s \(2^{\text {nd }}\) Law to find the acceleration of the block:
\(
\begin{aligned}
&\mathbf{R}=m \mathbf{a} \\
&\mathbf{a}=\frac{\mathbf{R}}{m} \\
&a=\frac{R}{m}
\end{aligned}
\)
\(
\begin{aligned}
a &=\frac{F \cos 20^{\circ}-F_f}{m} \\
a &=\frac{(600 \mathrm{~N})\left(\cos 20^{\circ}\right)-270 \mathrm{~N}}{50 \mathrm{~kg}} \\
a &=\frac{294 \mathrm{~N}}{50 \mathrm{~kg}} \\
a &=5.9 \mathrm{~N} / \mathrm{kg}=5.9 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
And with that we have found everything we needed:
\(
\begin{aligned}
&a=5.9 \mathrm{~m} / \mathrm{s}^2 \\
&F_{\mathrm{f}}=270 \mathrm{~N}
\end{aligned}
\)

Example 5.22: Hanna is pulling an object of \(20 \mathrm{~kg}\) over a horizontal plane. The force Hanna is exerting makes an angle of \(30^{\circ}\) with the horizontal. The coefficient of sliding friction \(\mu\), between the object and the plane, is \(0.57\). If the object is moving at a constant velocity, what is the magnitude of the force provided by Hanna?

Solution:

We will need to represent a horizontal surface, an object on it, and indicate that the object is moving at a constant velocity, as well as the fact that the object is being pulled by a force that makes \(30^{\circ}\) with the horizontal, and is subject to the force of friction:

By observing the sketch, we notice that the object is subject to 4 forces:

• Hanna’s pulling force, \(\mathbf{F}\)
• the sliding friction force, \(\mathbf{F}_{\mathrm{f}}\)
• the gravitational force, \(m \mathbf{g}\)
• and the normal force, \(\mathbf{N}\)
  So, the free-body diagram of the object will look something like this:

We know the angle that the pulling force makes with the horizontal \(\left(30^{\circ}\right)\), the mass of the object \((20 \mathrm{~kg})\), the coefficient of sliding friction (0.57), and that the object is moving at constant velocity. We need to find the magnitude of the pulling force exerted by Hanna.
We know
\(
\begin{aligned}
&\theta=30^{\circ} \\
&m=20 \mathrm{~kg} \\
&\mu=0.57
\end{aligned}
\)
\(\mathbf{v}\) is constant
We want to know
\(
F=\text { ? }
\)
The fact that the object is moving at constant velocity tells us that the object has no acceleration (because if it had, it wouldn’t be moving at constant velocity):
\(
\mathbf{a}=0
\)
And because the acceleration is zero, the resultant force acting on the object must also be zero (for Newton’s \(2^{\text {nd }}\) Law):
\(
\begin{aligned}
&\mathbf{R}=m \mathbf{a} \\
&\mathbf{R}=m \times 0 \\
&\mathbf{R}=0
\end{aligned}
\)
Knowing the resultant force, we can use the following strategy to find the magnitude of Hanna’s pull:

1. We find the \(x\) and \(y\) components of the resultant force, as a sum of the \(x\) and \(y\) components of all the forces that act on the object.
2. Because we already know that \(\mathbf{R}\) is zero, \(R_x\) and \(R_y\) must also be zero, so we substitute their values in the equations that we found in step 1.
3. We then use those equations to find \(F\) (the pulling force).

We begin by drawing coordinate axes on our free-body diagram and finding the components of all the forces that act on the object:

\(
\begin{array}{ll}
F_x=F \cos 30^{\circ} & F_y=F \sin 30^{\circ} \\
F_{\mathrm{f}_x}=-F_{\mathrm{f}} & F_{\mathrm{f}_y}=0 \\
N_x=0 & N_y=N \\
m g_x=0 & m g_y=-m g
\end{array}
\)
Next, we find the \(x\) and \(y\) components of the resultant force by adding all the \(x\) and \(y\) components:
\(x\) :
\(
\begin{aligned}
&R_x=F_x+F_{\mathrm{f}_x}+N_x+m g_x \\
&R_x=F \cos 30^{\circ}+\left(-F_{\mathrm{f}}\right)+0+0 \\
&R_x=F \cos 30^{\circ}-F_{\mathrm{f}}
\end{aligned}
\)
\(y\) :
\(
\begin{aligned}
&R_y=F_y+F_{\mathrm{f}_y}+N_y+m g_y \\
&R_y=F \sin 30^{\circ}+0+N+(-m g) \\
&R_y=F \sin 30^{\circ}+N-m g
\end{aligned}
\)
And since
\(
\begin{aligned}
&R_x=0 \\
&R_y=0
\end{aligned}
\)
we substitute their values and get
\(
\begin{aligned}
&0=F \cos 30^{\circ}-F_{\mathrm{f}} \quad \text { (1) } \\
&0=F \sin 30^{\circ}+N-m g \dots(2)
\end{aligned}
\)
These two equations have three unknowns \(\left(F, F_{\mathrm{f}}\right.\) and \(\left.N\right)\).
The number of equations has to be equal to the number of unknowns in order to solve them. Therefore, we should somehow reduce the number of unknowns to two.
We know the coefficient of sliding friction \(\mu\), and the sliding friction force has by definition the magnitude equal to \(\mu\) multiplied by \(N\) :
\(
F_{\mathrm{f}}=\mu N
\)
Which means that we have reduced the number of unknowns to two, and we can now solve the two equations.
By exchanging \(F_{\mathrm{f}}\) with \(\mu N\) in Eq. (1), we get
\(
0=F \cos 30^{\circ}-\mu N \dots(3)
\)
So, we now have two independent equations (Eq. (2) and Eq. (3)).
First, we solve one of them for one unknown: let’s solve Eq. (2) for \(N\) :

\(
\begin{aligned}
&0=F \sin 30^{\circ}+N-m g \\
&F \sin 30^{\circ}+N-m g=0 \\
&N=m g-F \sin 30^{\circ}
\end{aligned}
\)
And substitute \(N\) in Eq. (3):
\(
\begin{aligned}
&0=F \cos 30^{\circ}-\mu N \\
&0=F \cos 30^{\circ}-\mu\left(m g-F \sin 30^{\circ}\right)
\end{aligned}
\)
Finally, we solve this equation for \(F\) :
\(
\begin{aligned}
&0=F \cos 30^{\circ}-\mu\left(m g-F \sin 30^{\circ}\right) \\
&F \cos 30^{\circ}-\mu\left(m g-F \sin 30^{\circ}\right)=0 \\
&F \cos 30^{\circ}-\mu m g+\mu F \sin 30^{\circ}=0 \\
&F \cos 30^{\circ}+\mu F \sin 30^{\circ}=\mu m g \\
&F\left(\cos 30^{\circ}+\mu \sin 30^{\circ}\right)=\mu m g \\
&F=\frac{\mu m g}{\cos 30^{\circ}+\mu \sin 30^{\circ}} \\
&F=\frac{(0.57)(20 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})}{\cos 30^{\circ}+(0.57)\left(\sin 30^{\circ}\right)} \\
&F=\frac{112 \mathrm{~N}}{1.15} \\
&F=97 \mathrm{~N}
\end{aligned}
\)
Hence, Hanna’s pulling force has a magnitude of \(97 \mathrm{~N}\).

Example 5.23: Rob is pushing a block of \(12 \mathrm{~kg}\) up a frictionless ramp. The ramp makes an angle of \(35^{\circ}\) with the horizontal. Assuming that Rob exerts a force of \(148 \mathrm{~N}\), find:
a. The resultant force acting on the block.
b. The acceleration of the block.
c. The magnitude of the normal force.

Solution:

We need to represent a ramp that makes an angle of \(35^{\circ}\) with the horizontal, a block on it, and indicate that the block is pushed up by a force parallel to the ramp:

By looking at the sketch, we can conclude that there are three forces acting on the block:

• Rob’s push, \(\mathbf{F}\)
• the gravitational force, \(m \mathbf{g}\)
• and the normal force \(\mathbf{N}\), which the ramp exerts to prevent the penetration of the block
  Having figured out the forces acting on the block, let’s draw a free-body diagram of the block:

Let’s think about what we know, and what we’re asked to find:
We know the mass of the block \((12 \mathrm{~kg})\), the angle the ramp makes with the horizontal \(\left(35^{\circ}\right)\), and the force exerted by Rob \((148 \mathrm{~N})\).
We’re asked to find the resultant force acting on the block, the acceleration that the block has as a result, and the normal force exerted by the ramp on the block.
We know
\(
\begin{aligned}
&m=12 \mathrm{~kg} \\
&\theta=35^{\circ} \\
&F=148 \mathrm{~N}
\end{aligned}
\)
We want to know
\(
\begin{aligned}
&\mathbf{R}=\text { ? } \\
&\mathbf{a}=\text { ? } \\
&N=?
\end{aligned}
\)
Let’s start by finding the resultant force.
In this case we have a block moving up a ramp, so for our convenience, we will use tilted coordinate axes, with the \(x\) axis in the direction of motion (uphill).

After drawing the coordinate axes on the free-body diagram of the block, we proceed to find the components of the individual forces acting on the block:

Keep in mind that the angle that the gravitational force, \(m \mathbf{g}\), makes with its \(y\) component, \(m g_y\) is equal to the angle that the ramp makes with the horizontal ( \(35^{\circ}\) in this case).
\(\begin{array}{ll}F_x=F & F_y=0 \\ N_x=0 & N_y=N \\ m g_x=-m g \sin 35^{\circ} & m g_y=-m g \cos 35^{\circ}\end{array}\)
And the components of the resultant force will be:
\(x\):
\(
\begin{aligned}
&R_x=F_x+N_x+m g_x \\
&R_x=F+0+\left(-m g \sin 35^{\circ}\right) \\
&R_x=F-m g \sin 35^{\circ} \dots(1)
\end{aligned}
\)
\(y:\)
\(
\begin{aligned}
&R_y=F_y+N_y+m g_y \\
&R_y=0+N+\left(-m g \cos 35^{\circ}\right) \\
&R_y=N-m g \cos 35^{\circ} \dots(2)
\end{aligned}
\)
We know that the motion of the block is along the ramp, therefore \(R_y\) must be zero. Otherwise there would be an acceleration in the \(y\) direction, which is not the case.
\(
R_y=0
\)
Therefore, the magnitude of \(\mathbf{R}\) is equal to the absolute value of \(R_x\). We know that \(R_x\) must be positive because the block is accelerating in the positive \(x\) direction. So, we can write:
\(
R=R_x
\)
We have already determined \(R_x\) in Eq. (1):
\(
R_x=F-m g \sin 35^{\circ}
\)
Therefore,
\(
\begin{aligned}
&R=F-m g \sin 35^{\circ} \\
&R=148 \mathrm{~N}-(12 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})\left(\sin 35^{\circ}\right) \\
&R=148 \mathrm{~N}-67 \mathrm{~N} \\
&R=81 \mathrm{~N}
\end{aligned}
\)
So, the resultant force on the block is \(81 \mathrm{~N}\) and directed in the positive \(x\) direction:

Next, we need to find the acceleration.
Since we know the resultant force and the mass, we can apply Newton’s \(\mathbf{2}^{\text {nd }}\) Law to get the acceleration:
\(
\begin{aligned}
\mathbf{R} &=m \mathbf{a} \\
\mathbf{a} &=\frac{\mathbf{R}}{m} \\
a &=\frac{R}{m} \\
a &=\frac{81 \mathrm{~N}}{12 \mathrm{~kg}} \\
a &=6.8 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
Lastly, we need to find the magnitude of the normal force.
The normal force \(N\) appears in Eq. (2):
\(
R_y=N-m g \cos 35^{\circ}
\)
And we’ve shown that \(R_y\) is zero, therefore:
\(
\begin{aligned}
&0=N-m g \cos 35^{\circ} \\
&N=m g \cos 35^{\circ} \\
&N=(12 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})\left(\cos 35^{\circ}\right) \\
&N=96 \mathrm{~N}
\end{aligned}
\)
At this point, we’ve found everything we were asked to find:
\(
\begin{aligned}
&R=81 \mathrm{~N} \\
&a=6.8 \mathrm{~m} / \mathrm{s}^2 \\
&N=96 \mathrm{~N}
\end{aligned}
\)

Example 5.24: Micheal is pulling a mass of \(30 \mathrm{~kg}\) up an incline which makes an angle of \(26^{\circ}\) with the horizontal. The mass has an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^2\) and the coefficient of kinetic friction between the mass and the incline is \(0.58\). What is the magnitude of the pulling force Micheal is exerting?

Solution:

We draw an incline that makes an angle of \(26^{\circ}\) with the horizontal, a mass on it, and indicate that the mass is pulled upwards, has an acceleration directed upwards, and is subject to the force of friction:

Looking at the sketch, we can infer that 4 forces are acting on the mass:

• Micheal’s pull, \(\mathbf{F}\)
• the friction force, \(\mathbf{F}_{\mathrm{f}}\)
• the force due to gravity, \(m\) g
• and the normal force, \(\mathbf{N}\)
  Let’s draw a free-body diagram of the mass:

This is what we know: the mass \((30 \mathrm{~kg})\), the angle that the incline makes with the horizontal \(\left(26^{\circ}\right)\), the acceleration of the mass \(\left(2.5 \mathrm{~m} / \mathrm{s}^2\right)\), and the coefficient of kinetic friction (0.58).
We need to find the force exerted by Micheal.
We know
\(
\begin{aligned}
&m=30 \mathrm{~kg} \\
&\theta=26^{\circ} \\
&a=2.5 \mathrm{~m} / \mathrm{s}^2 \\
&\mu=0.58
\end{aligned}
\)
We want to know
\(
F=\text { ? }
\)
Since we know the mass and the acceleration, we can find the magnitude of the resultant force by applying Newton’s \(2^{\text {nd }}\) Law:
\(
\mathbf{R}=m \mathbf{a}
\)
Now that we determined the resultant force acting on the mass, we can find \(F\) using the following strategy:
1. We first find the \(x\) and \(y\) components of the resultant force, \(R_x\) and \(R_y\), as a sum of the \(x\) and \(y\) components of all the forces that act on the mass.
2. We then find the values of \(R_x\) and \(R_y\) by decomposing \(\mathbf{R}\), and substitute them in place of \(R_x\) and \(R_y\) in the equations that we found in the previous step.
3. Finally, we use the resulting equations to find \(F\).
Let’s begin with the first step of our strategy, by drawing the coordinate axes on our free-body diagram. For convenience, we choose the \(x\) axis to be in the direction of motion. Then, we determine the \(x\) and \(y\) components of all the forces that act on the mass:

\(
\begin{array}{ll}
F_x=F & F_y=0 \\
F_{\mathrm{f}_x}=-F_{\mathrm{f}} & F_{\mathrm{f}_y}=0 \\
N_x=0 & N_y=N \\
m g_x=-m g \sin 26^{\circ} & m g_y=-m g \cos 26^{\circ}
\end{array}
\)
And we find the components of the resultant force as a sum of the components of all the forces:
\(x\):
\(
\begin{aligned}
&R_x=F_x+F_{\mathrm{f}_x}+N_x+m g_x \\
&R_x=F+\left(-F_{\mathrm{f}}\right)+0+\left(-m g \sin 26^{\circ}\right) \\
&R_x=F-F_{\mathrm{f}}-m g \sin 26^{\circ} \dots(1)
\end{aligned}
\)
\(y\):
\(
\begin{aligned}
&R_y=F_y+F_{\mathrm{f}_y}+N_y+m g_y \\
&R_y=0+0+N+\left(-m g \cos 26^{\circ}\right) \\
&R_y=N-m g \cos 26^{\circ} \dots(2)
\end{aligned}
\)
The next step is to determine \(R_x\) and \(R_y\) by decomposing \(\mathbf{R}\) :
First of all, \(\mathbf{R}\) has the same direction as the acceleration of the mass, i.e. the positive \(x\) direction:

Therefore,
\(
\begin{aligned}
&R_x=R=m a \\
&R_y=0
\end{aligned}
\)
Now, let’s substitute these values in Eq. (1) and Eq. (2), respectively:
\(
\begin{aligned}
&R_x=F-F_{\mathrm{f}}-m g \sin 26^{\circ} \\
&R_y=N-m g \cos 26^{\circ} \\
&\downarrow \\
&m a=F-F_{\mathrm{f}}-m g \sin 26^{\circ} \dots(3)\\
&0=N-m g \cos 26^{\circ} \dots(4)
\end{aligned}
\)
We have two equations with 3 unknowns: \(F, F_f, N\).
To find \(F\), we need to reduce the number of unknowns to 2 .
We can do that by remembering that the magnitude of the friction force is:
\(
F_{\mathrm{f}}=\mu N
\)
We already know the value of the coefficient of kinetic friction \(\mu\). Therefore, Eq. (3) becomes:
\(
\begin{aligned}
&m a=F-F_{\mathrm{f}}-m g \sin 26^{\circ} \\
&m a=F-\mu N-m g \sin 26^{\circ} \dots(5)
\end{aligned}
\)
At this point, we have two equations, Eq. (4) and Eq. (5), with two unknowns, \(F\) and \(N\), so we can easily solve them.
Let’s solve Eq. (4) for \(N\) :
\(
\begin{aligned}
&0=N-m g \cos 26^{\circ} \\
&N-m g \cos 26^{\circ}=0 \\
&N=m g \cos 26^{\circ}
\end{aligned}
\)
And substitute it in Eq. (5):
\(
\begin{aligned}
&m a=F-\mu N-m g \sin 26^{\circ} \\
&m a=F-\mu m g \cos 26^{\circ}-m g \sin 26^{\circ} \\
&m a+\mu m g \cos 26^{\circ}+m g \sin 26^{\circ}=F \\
&F=m a+\mu m g \cos 26^{\circ}+m g \sin 26^{\circ} \\
&F=m\left(a+\mu g \cos 26^{\circ}+g \sin 26^{\circ}\right) \\
&F=(30 \mathrm{~kg})\left[2.5 \mathrm{~m} / \mathrm{s}^2+(0.58)\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)\left(\mathrm{cos} 26^{\circ}\right)+\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)\left(\sin 26^{\circ}\right)\right] \\
&F=(30 \mathrm{~kg})\left(2.5 \mathrm{~m} / \mathrm{s}^2+5.1 \mathrm{~m} / \mathrm{s}^2+4.3 \mathrm{~m} / \mathrm{s}^2\right) \\
&F=(30 \mathrm{~kg})\left(11.9 \mathrm{~m} / \mathrm{s}^2\right) \\
&F=360 \mathrm{~N}
\end{aligned}
\)
Hence, Micheal is pulling the mass with a force of \(360 \mathrm{~N}\).

Example 5.25: A mass of \(108 \mathrm{~g}\) is hanging from two massless ropes attached to the ceiling. One rope makes an angle of \(50^{\circ}\) with the ceiling, while the other makes an angle of \(29^{\circ}\). Find the tensions in the two ropes.

Solution:

Looking at our sketch, we can infer that the mass is subject to 3 forces:

• the tension force exerted by the first rope, \(\mathbf{T}_1\)
• the tension force exerted by the second rope, \(\mathbf{T}_2\)
• and the force of gravity, \(m \mathbf{g}\)
  Here’s the free-body diagram of our hanging mass:

We know the mass ( \(108 \mathrm{~g}\), which in kilograms is \(0.108 \mathrm{~kg})\), and the angles that the two ropes make with the ceiling \(\left(50^{\circ}\right.\) and \(\left.29^{\circ}\right)\).
We are asked to find the tensions in the two ropes.
We know
\(
\begin{aligned}
&m=0.108 \mathrm{~kg} \\
&\theta_1=50^{\circ} \\
&\theta_2=29^{\circ}
\end{aligned}
\)
We want to know
\(
\begin{aligned}
&T_1=\text { ? } \\
&T_2=\text { ? }
\end{aligned}
\)
The mass is hanging.
What does this tell us?
This tells us that the acceleration of the mass must be zero.
And because the acceleration is zero, the resultant force acting on the mass is also zero. Indeed, for Newton’s \(2^{\text {nd }}\) Law:
\(
\begin{aligned}
&\mathbf{R}=m \mathbf{a} \\
&\mathbf{R}=m \times 0 \\
&\mathbf{R}=0
\end{aligned}
\)

Now that we determined that the resultant force acting on the mass is zero, we can find the tensions of the two ropes using the following step-by-step process:
1. First, we find the \(x\) and \(y\) components of the resultant force, as a sum of the \(x\) and \(y\) components of all the forces that act on the mass.
2. Then, since we know that \(\mathbf{R}\) is zero, \(R_x\) and \(R_y\) must also be zero, so we substitute them with 0 in the equations that we found in the previous step.
3. Finally, we use the resulting equations to find the tensions, \(T_1\) and \(T_2\).
Let’s start with the first step.
We draw the coordinate axes on our free-body diagram. For convenience, we choose the \(x\) axis horizontal and the \(y\) axis vertical. Then, we determine the \(x\) and \(y\) components of all the forces that act on the mass.

We need to keep in mind that the angle between each tension force and its \(x\) component is equal to the angle that the rope, producing that tension, makes with the ceiling:

\(
\begin{array}{ll}
T_{1_x}=-T_1 \cos 50^{\circ} & T_{1_y}=T_1 \sin 50^{\circ} \\
T_{2_x}=T_2 \cos 29^{\circ} & T_{2_y}=T_2 \sin 29^{\circ} \\
m g_x=0 & m g_y=-m g
\end{array}
\)
Thus, the \(x\) and \(y\) components of the resultant force will be:
\(x:\)
\(
\begin{aligned}
&R_x=T_{1_x}+T_{2_x}+m g_x \\
&R_x=-T_1 \cos 50^{\circ}+T_2 \cos 29^{\circ}+0 \\
&R_x=T_2 \cos 29^{\circ}-T_1 \cos 50^{\circ} \dots(1)
\end{aligned}
\)
\(y:\)
\(
\begin{aligned}
&R_y=T_{1_y}+T_{2_y}+m g_y \\
&R_y=T_1 \sin 50^{\circ}+T_2 \sin 29^{\circ}+(-m g) \\
&R_y=T_1 \sin 50^{\circ}+T_2 \sin 29^{\circ}-m g \dots(2)
\end{aligned}
\)
The next step is to substitute \(R_x\) and \(R_y\) in Eq. (1) and Eq. (2) with 0:
\(
\begin{aligned}
&R_x=T_2 \cos 29^{\circ}-T_1 \cos 50^{\circ} \\
&R_y=T_1 \sin 50^{\circ}+T_2 \sin 29^{\circ}-m g
\end{aligned}
\)

\(
\begin{aligned}
&0=T_2 \cos 29^{\circ}-T_1 \cos 50^{\circ} \dots(3) \\
&0=T_1 \sin 50^{\circ}+T_2 \sin 29^{\circ}-m g \dots(4)
\end{aligned}
\)
Using these two equations we can easily find \(T_1\) and \(T_2\).
There are multiple ways in which we can do this. One way would be to first solve Eq. (3) for \(T_2\)
\(
\begin{aligned}
&0=T_2 \cos 29^{\circ}-T_1 \cos 50^{\circ} \\
&T_2 \cos 29^{\circ}-T_1 \cos 50^{\circ}=0 \\
&T_2 \cos 29^{\circ}=T_1 \cos 50^{\circ} \\
&T_2=T_1 \frac{\cos 50^{\circ}}{\cos 29^{\circ}} \\
&T_2=0.735 T_1 \dots(5)
\end{aligned}
\)
Then, we substitute \(T_2\) with \(0.735 T_1\) in Eq. (4):
\(
\begin{aligned}
&0=T_1 \sin 50^{\circ}+T_2 \sin 29^{\circ}-m g \\
&0=T_1 \sin 50^{\circ}+0.735 T_1 \sin 29^{\circ}-m g
\end{aligned}
\)
And we solve it for \(T_1\) :
\(
\begin{aligned}
&0=T_1\left(\sin 50^{\circ}+0.735 \sin 29^{\circ}\right)-m g \\
&0=T_1(1.12)-m g
\end{aligned}
\)
\(
\begin{aligned}
&m g=1.12 T_1 \\
&1.12 T_1=m g \\
&T_1=\frac{m g}{1.12} \\
&T_1=\frac{(0.108 \mathrm{~kg})(9.81 \mathrm{~N} / \mathrm{kg})}{1.12} \\
&T_1=0.946 \mathrm{~N}
\end{aligned}
\)
Finally, we substitute the value of \(T_1\) in Eq. (5) to find \(T_2\) :
\(
\begin{aligned}
&T_2=0.735 T_1 \\
&T_2=(0.735)(0.946 \mathrm{~N}) \\
&T_2=0.695 \mathrm{~N}
\end{aligned}
\)
Therefore, the tension in the rope at the \(50^{\circ}\) angle is \(0.946 \mathrm{~N}\), and the tension in the rope at the \(29^{\circ}\) angle is \(0.695 \mathrm{~N}\).

Example 5.26: We have two objects – a cube of \(13 \mathrm{~kg}\) and a sphere of \(39 \mathrm{~kg}\). The cube hangs from a rope attached to the ceiling, while the sphere hangs from a second rope attached to the bottom of the cube. Assuming the two ropes are massless, What is the tension in the first rope? And in the second?

Solution: 

Let’s start by drawing a sketch of what is happening:

Since we’re dealing with massless ropes, we need to keep in mind that tensions exerted by the ends of a massless rope are equal in magnitude. We will indicate the magnitude of the tensions in the first rope with \(T_1\), and the magnitude of the tensions in the second rope with \(T_2\).

Let’s carefully examine our sketch and enumerate all the forces that we think act on our two objects:
The cube is subject to 3 forces:

• the tension in the upper rope, \(\mathbf{T}_1\) (which prevents the cube from falling)
• the tension in the lower rope, \(\mathbf{T}_2\) (pulling the cube downward)
• and the gravitational force, \(m \mathbf{g}\)

The sphere is subject to only 2 forces:

• the tension in the lower rope, \(\mathbf{T}_2\) (which prevents the sphere from falling)
• and the gravitational force, \(M \mathbf{g}\)
  Let’s draw a free-body diagram for the cube, and another for the sphere:

We know the masses ( \(13 \mathrm{~kg}\) for the cube and \(39 \mathrm{~kg}\) for the sphere).
We want to find the tensions in the two ropes.
We know
\(
\begin{aligned}
&m=13 \mathrm{~kg} \\
&M=39 \mathrm{~kg}
\end{aligned}
\)
We want to know
\(
\begin{aligned}
&T_1=? \\
&T_2=?
\end{aligned}
\)
The cube and the sphere are hanging, i.e. they are in static equilibrium. This means that the resultant forces on the cube and on the sphere must be zero.
The sphere is subject to two forces that are opposite in direction ( \(\mathbf{T}_2\) and \(\left.M \mathbf{g}\right)\). Since the resultant force is zero, these two forces must be equal in magnitude:
\(
\begin{aligned}
&T_2=M g \\
&T_2=(39 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg}) \\
&T_2=380 \mathrm{~N}
\end{aligned}
\)
The cube is subject to three parallel forces \(\left(\mathbf{T}_1, \mathbf{T}_2\right.\), and \(\left.m \mathbf{g}\right) . \mathbf{T}_1\) is directed upward, \(\mathbf{T}_2\) and \(m \mathbf{g}\) are directed downward. Again, since the resultant force is zero, the magnitude of \(\mathbf{T}_1\) must be equal to the sum of the magnitudes of \(\mathbf{T}_2\) and \(m \mathbf{g}\) :
\(
\begin{aligned}
&T_1=T_2+m g \\
&T_1=380 \mathrm{~N}+(13 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg}) \\
&T_1=510 \mathrm{~N}
\end{aligned}
\)
Hence, the tension in the upper rope is \(510 \mathrm{~N}\), and the tension in the lower rope is \(380 \mathrm{~N}\).

Example 5.27: Two masses of \(80 \mathrm{~kg}\) and \(140 \mathrm{~kg}\) hang from a rope that runs over a pulley. You can assume that the rope is massless and inextensible and that the pulley is frictionless. Find the upward acceleration of the smaller mass and the tension in the rope.

Solution:

Let’s start by drawing a sketch of what is happening:

We have a massless rope that runs over a frictionless pulley, this means that the two masses are subject to upward tensions equal in magnitude. We will indicate the magnitude of the tensions with \(T\). In the case of the smaller mass, the tension wins the force of gravity, which means that the smaller mass is accelerating upward. However, in the case of the larger mass, the force of gravity wins the tension, which means that the larger mass is accelerating downward.

Also, since the rope is inextensible, the two masses move with accelerations that are equal in magnitude. We will indicate the magnitude of the accelerations with \(a\). With all of that said, let’s list all the forces that act on the two masses.
The smaller mass is subject to 2 forces:

• the upward tension, \(\mathbf{T}\)
• and the force of gravity, \(m \mathbf{g}\)
  The larger mass is also subject to 2 forces:
• the upward tension, \(\mathbf{T}\)
• and the force of gravity, \(M \mathbf{g}\)
  Here are the free-body diagrams of the two masses:

We know the smaller mass \((80 \mathrm{~kg})\) and the larger mass \((140 \mathrm{~kg})\).
We want to find the acceleration of the smaller mass (which, as we saw has the same magnitude as the acceleration of the larger mass), and the tension in the rope.
We know
\(
\begin{aligned}
&m=80 \mathrm{~kg} \\
&M=140 \mathrm{~kg}
\end{aligned}
\)
We want to know
\(
\begin{aligned}
&a=? \\
&T=?
\end{aligned}
\)
The smaller mass is subject to two forces: \(\mathbf{T}\) (directed upward) and \(m \mathbf{g}\) (directed downward), where \(\mathbf{T}\) is larger in magnitude. Therefore, the resultant force \((\mathbf{r})\) will be directed upward, and have the magnitude equal to the difference between \(T\) and \(m g\) :
\(
r=T-m g \quad \text { (1) }
\)
The larger mass is also subject to two forces: \(\mathbf{T}\) (directed upward) and \(M \mathbf{g}\) (directed downward), where \(M g\) is larger in magnitude. Therefore, the resultant force \((\mathbf{R})\) will be directed downward, and have the magnitude equal to the difference between \(M g\) and \(T\) :
\(
R=M g-T \quad \text { (2) }
\)
We now have two equations with 3 unknowns: \(r, R, T\).

We can reduce the number of unknowns to 2 by remembering that the two masses have accelerations equal in magnitude. Indeed, applying Newton’s \(2^{\text {nd }}\) Law, the magnitudes of the two resultant forces can be expressed as:
\(
\begin{aligned}
&r=m a \\
&R=M a
\end{aligned}
\)
So, we can substitute \(r\) and \(R\) in Eq. (1) and Eq. (2):
\(
\begin{aligned}
&r=T-m g \\
&R=M g-T \\
&\downarrow \\
&m a=T-m g \dots(3) \\
&M a=M g-T \dots(4)
\end{aligned}
\)
We now have two equations with 2 unknowns ( \(a\) and \(T\) ), so we can solve them.
Let’s solve Eq. (3) for \(T\) :
\(
\begin{aligned}
&m a=T-m g \\
&m a+m g=T \\
&T=m a+m g \dots(5)
\end{aligned}
\)
Next, let’s substitute \(T\) with \(m a+m g\) in Eq. (4):
\(
\begin{aligned}
&M a=M g-T \\
&M a=M g-(m a+m g) \\
&M a=M g-m a-m g
\end{aligned}
\)

And solve this equation for \(a\) :
\(
\begin{aligned}
&M a+m a=M g-m g \\
&a(M+m)=g(M-m) \\
&a=\frac{M-m}{M+m} g \\
&a=\frac{140 \mathrm{~kg}-80 \mathrm{~kg}}{140 \mathrm{~kg}+80 \mathrm{~kg}}\left(9.8 \mathrm{~m} / \mathrm{s}^2\right) \\
&a=\frac{60 \mathrm{~kg}}{220 \mathrm{~kg}}\left(9.8 \mathrm{~m} / \mathrm{s}^2\right) \\
&a=2.7 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
Finally, we can find the magnitude of the tension \(T\) using Eq. (5):
\(
\begin{aligned}
&T=m a+m g \\
&T=m(a+g) \\
&T=(80 \mathrm{~kg})\left(2.7 \mathrm{~m} / \mathrm{s}^2+9.8 \mathrm{~m} / \mathrm{s}^2\right) \\
&T=1.0 \times 10^3 \mathrm{~N}
\end{aligned}
\)
Therefore, the smaller mass has an acceleration of \(2.7 \mathrm{~m} / \mathrm{s}^2\) (which is also the magnitude of the acceleration of the larger mass), and the tension in the rope is \(1.0 \times 10^3 \mathrm{~N}\).

Example 5.28: Two wooden blocks, connected by a massless string, are moving at a constant velocity over a floor with friction. The smaller of the two blocks is pulled by a horizontal force of \(500 \mathrm{~N}\). Knowing that the mass of the smaller block is \(30 \mathrm{~kg}\) and the mass of the larger block is \(40 \mathrm{~kg}\), find the tension in the string and the coefficient of sliding friction between the blocks and the floor.

Solution:

Let’s begin by drawing a sketch that represents what is happening in the problem. In the sketch, we draw a horizontal floor with two blocks that are connected by a string sitting on it. Then, we indicate that the smaller of the two blocks is pulled by a horizontal force and that both the blocks are subject to the force of friction. It’s also a good idea to write that the two blocks move at constant velocity. The final sketch looks something like this:

Since the string connecting the two blocks is massless, the tension forces exerted by the ends of the string have the same magnitude, which we will indicate with \(T\).

Also, since the two blocks are made of the same material (wood) and they move on the same floor, the coefficient of sliding friction is the same for both blocks, and we will indicate it with \(\mu\).

In this problem, we are dealing with two blocks, so it’s a good idea to analyze them separately.
Looking carefully at our sketch, we can see that the smaller block is subject to 5 forces:

• The pulling force, \(\mathbf{F}\)
• The tension force, \(\mathbf{T}\)
• The friction force, \(\mathbf{f}_{\mathrm{f}}\)
• The gravitational force, \(m \mathbf{g}\)
• The normal force, \(\mathbf{n}\)
  Here’s the free-body diagram of the smaller block:

The larger block, on the other hand, is subject to 4 forces:

• The tension force, \(\mathbf{T}\)
• The friction force, \(\mathbf{F}_{\mathrm{f}}\)
• The gravitational force, \(M \mathbf{g}\)
• The normal force, \(\mathbf{N}\)
  Here’s the free-body diagram of the larger block:

We know the masses of the two blocks ( \(30 \mathrm{~kg}\) and \(40 \mathrm{~kg}\) ), the magnitude of the pulling force \((500 \mathrm{~N})\), and that the two blocks are moving at constant velocity.

We need to find the magnitude of the tension in the string \((T)\), and the coefficient of sliding friction \((\mu)\).
We know
\(
\begin{aligned}
&m=30 \mathrm{~kg} \\
&M=40 \mathrm{~kg} \\
&F=500 \mathrm{~N}
\end{aligned}
\)
\(\mathbf{v}\) is constant
We want to know
\(
T=\text { ? }
\)
\(
\mu=?
\)
Since the blocks are moving at a constant velocity, they have zero acceleration, so the net force on the smaller block \(\left(\mathbf{f}_{\text {net }}\right)\) is zero, and the net force on the larger block \(\left(\mathbf{F}_{\text {net }}\right)\) is also zero:
\(
\begin{aligned}
&\mathbf{f}_{\text {net }}=m \mathbf{a}=m \times 0=0 \\
&\mathbf{F}_{\text {net }}=M \mathbf{a}=M \times 0=0
\end{aligned}
\)

We can find the tension in the string and the coefficient of sliding friction following these steps:

1. First, we find the \(x\) and \(y\) components of the net force that acts on the smaller block in terms of the \(x\) and \(y\) components of all the forces that act on the smaller block.
2. Then, since \(\mathbf{f}_{\text {net }}\) is zero, \(f_{\text {net }_x}\) and \(f_{\text {net }_y}\) must also be zero, so we replace them with 0 in the equations that we found in step 1, and arrive at a final equation.
3. We repeat the equivalent of the previous steps for the larger block and arrive at a final equation for the larger block.
4. Lastly, we solve the two final equations and find the tension \(T\) and the coefficient of sliding friction \(\mu\).
Let’s begin with step 1.

To find the components of the net force that acts on the smaller block, we first draw the coordinate axes on the free-body diagram of the smaller block and enumerate the components of all the forces that act on it:

\(
\begin{array}{ll}
F_x=F & F_y=0 \\
T_x=-T & T_y=0 \\
f_{\mathrm{f}_x}=-f_{\mathrm{f}} & f_{\mathrm{f}_y}=0 \\
n_x=0 & n_y=n \\
m g_x=0 & m g_y=-m g
\end{array}
\)
We can now find the \(x\) and \(y\) components of the net force by adding the \(x\) and \(y\) components of all the forces:
\(x\) :
\(
\begin{aligned}
&f_{\text {net }_x}=F_x+T_x+f_{\mathrm{f}_x}+n_x+m g_x \\
&f_{\text {net }_x}=F+(-T)+\left(-f_{\mathrm{f}}\right)+0+0 \\
&f_{\text {net }_x}=F-T-f_{\mathrm{f}}
\end{aligned}
\)
\(y\) :
\(
\begin{aligned}
&f_{\text {net }_y}=F_y+T_y+f_{\mathrm{f}_y}+n_y+m g_y \\
&f_{\text {net }_y}=0+0+0+n+(-m g) \\
&f_{\text {net }_y}=n-m g
\end{aligned}
\)
And we replace \(f_{\text {net }_x}\) and \(f_{\text {net }_y}\) with 0 :
\(
\begin{aligned}
&0=F-T-f_{\mathrm{f}} \dots(1) \\
&0=n-m g \dots(2)
\end{aligned}
\)
Additionally, we know that the magnitude of the friction force \(\mathbf{f}_{\mathrm{f}}\) is the product between the coefficient of friction \(\mu\) and the magnitude of the normal force \(\mathbf{n}\) :
\(
f_{\mathrm{f}}=\mu n
\)
Therefore, Eq. (1) becomes:
\(
0=F-T-\mu n \dots(3)
\)
Next, let’s solve Eq. (2) for \(n\) :
\(
\begin{aligned}
&0=n-m g \\
&n-m g=0 \\
&n=m g
\end{aligned}
\)
And replace \(n\) with \(m g\) in Eq. (3):
\(
0=F-T-\mu m g \dots(4)
\)
This final equation has 2 unknowns \((T\) and \(\mu)\).
Let’s now repeat the same process on the larger block in order to find the second equation.
Again, we draw the coordinate axes on the free-body diagram and determine the \(x\) and \(y\) components of each force:

\(
\begin{array}{ll}
T_x=T & T_y=0 \\
F_{\mathrm{f}_x}=-F_{\mathrm{f}} & F_{\mathrm{f}_y}=0 \\
N_x=0 & N_y=N \\
M g_x=0 & M g_y=-M g
\end{array}
\)
Then, we calculate the \(x\) and \(y\) components of the net force \(\mathbf{F}_{\text {net }}\) :
\(x\) :
\(
\begin{aligned}
&F_{\text {net }_x}=T_x+F_{\mathrm{f}_x}+N_x+M g_x \\
&F_{\text {net }_x}=T+\left(-F_{\mathrm{f}}\right)+0+0 \\
&F_{\text {net }_x}=T-F_{\mathrm{f}}
\end{aligned}
\)
\(y:\)
\(
\begin{aligned}
&F_{\text {net }_y}=T_y+F_{\mathrm{f}_y}+N_y+M g_y \\
&F_{\text {net }_y}=0+0+N+(-M g) \\
&F_{\text {net }_y}=N-M g
\end{aligned}
\)
Again, because \(\mathbf{F}_{\text {net }}\) is zero, the components are zero as well, so we can replace them in the above equations with 0 :
\(
\begin{aligned}
&0=T-F_{\mathrm{f}} \dots(5) \\
&0=N-M g \dots(6)
\end{aligned}
\)
We can write the magnitude of the friction force as:
\(
F_{\mathrm{f}}=\mu N
\)
Therefore, Eq. (5) becomes:
\(
0=T-\mu N \quad \text { (7) }
\)
Next, let’s solve Eq. (6) for \(N\) :
\(
\begin{aligned}
&0=N-M g \\
&N-M g=0 \\
&N=M g
\end{aligned}
\)
And swap \(N\) with \(M g\) in Eq. (7):
\(
0=T-\mu M g \dots(8)
\)
This is the final equation for the larger block.
Thus, the two final equations are Eq. (4) and Eq. (8):
\(
\begin{aligned}
&0=F-T-\mu m g \\
&0=T-\mu M g
\end{aligned}
\)
These are two equations with two unknowns ( \(T\) and \(\mu)\), so we can solve them.
Let’s solve Eq. (8) for \(T\) :

\(
\begin{aligned}
&0=T-\mu M g \\
&T-\mu M g=0 \\
&T=\mu M g \dots(9)
\end{aligned}
\)
And replace \(T\) with \(\mu M g\) in Eq. (4):
\(
\begin{aligned}
&0=F-T-\mu m g \\
&0=F-\mu M g-\mu m g
\end{aligned}
\)
Now, let’s solve this equation for \(\mu\) :
\(
\begin{aligned}
&0=F-\mu M g-\mu m g \\
&\mu M g+\mu m g=F \\
&\mu g(M+m)=F \\
&\mu=\frac{F}{g(M+m)} \\
&\mu=\frac{500 \mathrm{~N}}{(9.8 \mathrm{~N} / \mathrm{kg})(40 \mathrm{~kg}+30 \mathrm{~kg})} \\
&\mu=\frac{500 \mathrm{~N}}{(9.8 \mathrm{~N} / \mathrm{kg})(70 \mathrm{~kg})} \\
&\mu=0.73
\end{aligned}
\)
Lastly, we can get \(T\) from Eq. (9):
\(
T=\mu M g
\)
\(
\begin{aligned}
&T=(0.73)(40 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg}) \\
&T=2.9 \times 10^2 \mathrm{~N}
\end{aligned}
\)
Therefore, the tension in the string is \(2.9 \times 10^2 \mathrm{~N}\), and the coefficient of sliding friction is \(0.73\).