Laws of Motion Quiz-L1

Hint

The acceleration of the electron, \(a=\frac{F}{m}=\frac{10^{-25}}{10^{-30}}=10^5 \mathrm{~ms}^{-2}\)
The time taken by the electron \((t)\) to cover the distance \((s)\) of \(0.2 \mathrm{~m}\) can be given by
\(
\begin{array}{rlrl}
s & =u t+\frac{1}{2} a t^2 \\
\Rightarrow 0.2 & =0+\frac{1}{2} \times 10^5 \times t^2 \\
\Rightarrow t^2 & =0.4 \times 10^{-5}=4 \times 10^{-6} \Rightarrow t=2 \times 10^{-3} \mathrm{~s}
\end{array}
\)

Hint

The retardation \(a\) of the bullet is given by
\(
\begin{array}{r}
a=-\frac{u^2}{2 s}=\frac{-90 \times 90}{2 \times 0.6} \quad\left(\text { From, } v^2=u^2+2 a s, v=0\right) \\
\left(\because \text { Given, } s=60 \mathrm{~cm}=0.6 \mathrm{~m}, u=90 \mathrm{~ms}^{-1}\right)
\end{array}
\)
\(
\Rightarrow \quad a=-6750 \mathrm{~ms}^{-2}
\)

From second law of motion,
Retarding force, \(F=m a=0.06 \times 6750\)
\(
\Rightarrow \quad F=405 \mathrm{~N}
\)

Hint

Given, \(u=20 \mathrm{~ms}^{-1}, v=0, s=50 \mathrm{~m}\) and \(m=1 \mathrm{~kg}\)
To calculate force, we have the formula \(F=m a\), but we have to first calculate acceleration \(a\).
Using the third equation of motion, i.e.
\(
\begin{array}{rlrl}
v^2 & =u^2+2 a s \\
& & (0)^2 & =(20)^2+2 \times a \times 50 \\
\Rightarrow & 100 a & =-400 \\
\Rightarrow & & a & =-4 \mathrm{~ms}^{-2}
\end{array}
\)

Acceleration \(a=-4 \mathrm{~ms}^{-2}\) (-ve sign shows that speed of the stone decreases, i.e. retardation)
Now,
\(
\begin{aligned}
F & =m a=(1 \mathrm{~kg}) \times\left(-4 \mathrm{~ms}^{-2}\right) \\
& =-4 \mathrm{~kg}-\mathrm{ms}^{-2}=-4 \mathrm{~N}
\end{aligned}
\)

Thus, force of opposition between the stone and the ice is \(-4 \mathrm{~N}\). The negative value of force shows that the opposing force acts in a direction opposite to the direction of motion.

Hint

Force exerting on block, \(F=v \frac{d m}{d t}=4 \times 3=12 \mathrm{~N}\)
So, acceleration of the block, \(a=\frac{F}{m}=\frac{12}{6}=2.0 \mathrm{~ms}^{-2}\)

Hint

Free body diagram is shown below.

where, \(F_x=\) horizontal component of the force and \(\quad F_y=\) vertical component of the force.
Horizontal component of the force \(=50 \sin 60^{\circ}=\frac{50 \sqrt{3}}{2} \mathrm{~N}\) Acceleration of the block, \(a\)
\(
\begin{aligned}
& =\frac{\text { Component of force in the direction of acceleration }}{\text { Mass }} \\
& =\frac{50 \sqrt{3}}{2} \times \frac{1}{5}=5 \sqrt{3} \mathrm{~ms}^{-2}
\end{aligned}
\)

Hint

\(\because\) Acceleration, \(a=\frac{|\mathrm{F}|}{m}\)
\(
\Rightarrow \text { Mass, } m=\frac{|\mathbf{F}|}{a}=\frac{\sqrt{6^2+8^2+10^2}}{\sqrt{2}}=10 \mathrm{~kg}
\)

Hint

Given, \(\mathbf{F}=\left(2 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) N \Rightarrow \frac{d \mathbf{p}}{d t}=2 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\)
\(
\begin{array}{ll}
\Rightarrow & d \mathbf{p}=2 t d t \hat{\mathbf{i}}+3 t^2 d \hat{\mathbf{j}} \\
\Rightarrow & \int d \mathbf{p}=2 \int_0^2 t d t \hat{\mathbf{i}}+3 \int_0^2 t^2 d t \hat{\mathbf{j}} \\
\Rightarrow & \Delta \mathbf{p}=\left[t^2\right]_0^2 \hat{\mathbf{i}}+\left[t^3\right]_0^2 \hat{\mathbf{j}} \\
\Rightarrow & \Delta \mathbf{p}=4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}} \\
\Rightarrow & |\Delta \mathbf{p}|=\sqrt{16+64}=\sqrt{80} \approx 9 \mathrm{~kg} \mathrm{~ms}^{-1}
\end{array}
\)

Hint

Unbalanced external force, \(F=F_1-F_2=10-2=8 \mathrm{~N}\)
So,\(F=m a\)
\(\Rightarrow\) Acceleration,
\(
a=\frac{F}{m}=\frac{8}{2}=4 \mathrm{~ms}^{-2}
\)

The body moves in direction of \(F_1\)

Hint

Given, \(m=0.15 \mathrm{~kg}, v=12 \mathrm{~ms}^{-1}\) and \(u=-12 \mathrm{~ms}^{-1}\) Change in momentum,
\(
\begin{gathered}
p_2-p_1=m(v-u)=0.15[12-(-12)]=0.15 \times 24 \\
p_2-p_1=3.60 \mathrm{~kg}-\mathrm{ms}^{-1} \\
\text { Impulse, } I=p_2-p_1 \Rightarrow I=3.6 \mathrm{~N}-\mathrm{s}
\end{gathered}
\)

Hint

(i) Impulse \(=F \times t=m(v-u)=1(0-6)=-6 \mathrm{~N}-\mathrm{s}\)
(ii) Average retarding force that stops the hammer,
\(
F=\frac{\text { Impulse }}{\text { Time }}=\frac{6}{0.1}=60 \mathrm{~N}
\)
(iii) Average retardation, \(a=\frac{F}{m}=\frac{60}{1}=60 \mathrm{~ms}^{-2}\)

Hint

According to given question, change in momentum of the ball,
\(
\Delta p=p_f-p_i=m(v-u)=150 \times 10^{-3}[20-(-12)]=4.8 \mathrm{~N}-\mathrm{s}
\)

So, by impulse-momentum theorem, impulse, \(I=\Delta p=4.8 \mathrm{~N}-\mathrm{s}\) and by time averaged definition of force in case of impulse
\(
\Rightarrow \quad F_{\mathrm{av}}=\frac{I}{\Delta t}=\frac{\Delta p}{\Delta t}=\frac{4.80}{0.01}=480 \mathrm{~N}
\)

Hint

(i) Impulse \(=\) Area under \(F\) – \(t\) curve
\(
\begin{aligned}
& =\text { Area of } \triangle A B C=\frac{1}{2} \times O P \times A C=\frac{1}{2} \times 18000 \times(2.5-1) \\
& =1.35 \times 10^4 \mathrm{~kg}-\mathrm{ms}^{-1} \text { or } \mathrm{N}-\mathrm{s}
\end{aligned}
\)
(ii) Average force \(=\frac{\text { Impulse }}{\text { Time }}=\frac{1.35 \times 10^4}{(2.5-1)}=9000 \mathrm{~N}\)

Hint

In mode (a), the man applies a force equal to \(25 \mathrm{~kg}\) weight in upward direction. According to Newton’s third law of motion, there will be a downward force of reaction on the floor.
\(\therefore\) Total action on the floor by the man
\(
\begin{aligned}
& =50 \mathrm{~kg}-\mathrm{wt}+25 \mathrm{~kg}-\mathrm{wt} \\
& =75 \mathrm{~kg}-\mathrm{wt} \\
& =75 \times 9.8 \mathrm{~N}=735 \mathrm{~N}
\end{aligned}
\)
In mode (b), the man applies a downward force on rope equal to \(25 \mathrm{~kg}\)-wt. According to Newton’s third law, the reaction will be in the upward direction by the rope on the man, so he becomes light by \(25 \mathrm{~kg}\)-wt.
\(\therefore\) Total action on the floor by the man
\(
\begin{aligned}
& =50 \mathrm{~kg} \text {-wt}-25 \mathrm{~kg} \text {-wt } \\
& =25 \mathrm{~kg} \text {-wt } \\
& =25 \times 9.8 \mathrm{~N}=245 \mathrm{~N}
\end{aligned}
\)

As the floor yields to a downward force of \(700 \mathrm{~N}\), so the man should adopt mode (b).

Hint

From the law of conservation of momentum,
\(
m_G v_G=m_B v_B
\)
where, \(m_G, v_G=\) mass and velocity of gun
\(m_B, v_B=\) mass and velocity of bullet
\(
\Rightarrow \quad v_B=\frac{m_G v_G}{m_B}=\frac{1 \times 5}{10 \times 10^{-3}}=500 \mathrm{~ms}^{-1}
\)

Hint

From the law of conservation of momentum,
\(
\begin{aligned}
p_3 & =\sqrt{p^2+p_2^2} \\
\Rightarrow \quad m_3 \times 4 & =\sqrt{(1 \times 12)^2+(2 \times 8)^2}=20 \\
\Rightarrow \quad m_3 & =5 \mathrm{~kg}
\end{aligned}
\)

Hint

Given,
\(
\begin{aligned}
m_1 & =m_2=5 \mathrm{~kg}, \\
u_1 & =3 \mathrm{~m} / \mathrm{s}, u_2=-3 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Before collision,

Let the velocity of the combined object be \(v\). Then, after collision,

Total momentum of the system before collision is
\(
m_1 u_1+m_2 u_2=5 \times 3+5 \times(-3)=0
\)

Total momentum of the system after collision is
\(
m_1 v+m_2 v=\left(m_1+m_2\right) v=(5+5) v=10 v
\)

According to the law of conservation of momentum, Momentum before collision \(=\) Momentum after collision
\(
\therefore, 0=10 v \Rightarrow v=0
\)

Hence, the velocity of the combined object after collision is zero.

Hint

According to the question,
\(
\begin{aligned}
& \text { (i) } \mathbf{p}_i=m v \sin 30^{\circ} \hat{\mathbf{i}}-m v \cos 30^{\circ} \hat{\mathbf{j}}, \mathbf{p}_f=-m v \sin 30^{\circ} \hat{\mathbf{i}}-m v \cos 30^{\circ} \hat{\mathbf{j}} \\
& \therefore \text { Impulse }=\Delta \mathbf{p}=\mathbf{p}_f-\mathbf{p}_i=-2 m v \sin 30^{\circ} \hat{\mathbf{i}}=-m v \hat{\mathbf{i}} \\
& \text { Magnitude of impulse }=|\Delta \mathbf{p}|=m v
\end{aligned}
\)

(ii) Negative sign of the impulse shows that it is along negative \(x\)-direction. Since, impulse and force are in the same direction, the force on the ball is along the negative direction of \(X\)-axis. Hence, the force on the wall will be along positive \(X\)-axis.

Hint

1. According to Newton’s first law of motion, “A body continues to be in its state of rest or uniform motion along a given direction, until and unless an external force is applied on it.” This is also known as the law of inertia.
2. Inertia is the property of a body by which it is unable to change its position of rest or uniform motion.
3. The mass of a body can be defined as the measure of inertia of a body.
4. Thus, inertia depends upon mass. The greater the mass of a body, the greater will be the inertia and the more difficult it will be to move the body.

Hint

(d) Given, mass, \(m=6 \mathrm{~kg}\)
Velocity, \(\Delta v=v_2-v_1=5-3=2 \mathrm{~ms}^{-1}\)
\(\therefore\) Momentum, \(\Delta p=m \Delta v=6 \times 2=12 \mathrm{~N}-\mathrm{s}\)

Hint

(a) Given, \(p=2+3 t^2\)

Differentiate w.r.t. \(t\), we get
\(
\frac{d p}{d t}=0+3 \times 2 t=6 t
\)

If \(t=3 \mathrm{~s}\), then \(\frac{d p}{d t}=6 \times 3=18 \mathrm{~N}\)
\(\therefore\) Force acting on the particle \(=18 \mathrm{~N}\)

Hint

If a body is acted upon by balanced forces then the net force acting on the body is zero.
We know from Newton’s law of motion \(\mathrm{F}=\mathrm{ma}\)
If \(\mathrm{F}=0\) then \(\mathrm{a}=0\)
Thus, the body will not accelerate. It will either be at rest or move with a constant velocity.

Hint

(b) Given,
\(
\begin{aligned}
F & =72 \text { dyne } \Rightarrow F=72 \times 10^{-5} \mathrm{~N}, \\
\theta & =60^{\circ}, m=9 \mathrm{~g}=9 \times 10^{-3} \mathrm{~kg} \\
F^{\prime} & =m a
\end{aligned}
\)

In this case, \(F^{\prime}=F \cos 60^{\circ}\)
\(
\begin{aligned}
\Rightarrow \quad a & =\frac{F^{\prime}}{m}=\frac{F \cos 60^{\circ}}{9 \times 10^{-3}} \\
& =\frac{72 \times 10^{-5} \times \cos 60^{\circ}}{9 \times 10^{-3}}=8 \times 10^{-2} \times \frac{1}{2} \\
& =4 \times 10^{-2} \mathrm{~ms}^{-1}=4 \mathrm{cms}^{-1}
\end{aligned}
\)

Hint

(d) Given, \(m=60 \mathrm{~kg}\) and \(a=1 \mathrm{~ms}^{-2}\)
\(\therefore\) Net force, \(F_{\text {net }}=\) Mass \(\times\) Acceleration \(=60 \times 1=60 \mathrm{~N}\)

Hint

(a) Given, \(m=3 \mathrm{~kg}, \Delta v=35-2=15 \mathrm{~ms}^{-1}\) and \(\Delta t=25 \mathrm{~s}\)

Force, \(F=m a=m \frac{\Delta v}{\Delta t}=3 \times \frac{1.5}{25}=0.18 \mathrm{~N}\), in the direction of motion

Hint

(b) Given, \(m=5 \mathrm{~kg}, F_1=8 \mathrm{~N}\) and \(F_2=6 \mathrm{~N}\)
Resultant force, \(F=\sqrt{8^2+6^2}=10 \mathrm{~N}\)
\(
\therefore a=\frac{F}{m}=\frac{10}{5}=2 \mathrm{~ms}^{-2}
\)
and \(\theta=\cos ^{-1}\left(\frac{6}{10}\right)=\cos ^{-1}(0.6)\), from \(6 \mathrm{~N}\)

Hint

Impulse = change in momentum
\(
=m { (v_f-v_i) }
\)
\(
\begin{aligned}
v_f & =0 \\
& =m(0-\vec{v}) \\
& =-m \vec{v}
\end{aligned}
\)

Hint

(a) Given, \(F=50 \mathrm{~N}, m=20 \mathrm{~kg}\) and \(v=15 \mathrm{~ms}^{-1}\)

Impulse \(=F \Delta t=m v\)
\(\therefore\) Time, \(\Delta t=\frac{m v}{F}=\frac{20 \times 15}{50}=6 \mathrm{~s}\)

Hint

\(
\begin{aligned}
& v^2=u^2+2 g h \\
& v^2=0+2 \times 9.8 \times 9.8
\end{aligned}
\)
\(
\begin{aligned}
v^2 & =\sqrt{2 \times 9.8 \times 9.8} \\
\Rightarrow \quad v & =9.8 \sqrt{2} \mathrm{~m} / \mathrm{s} \\
\text { Again, } v^2 & =u^2-2 g h \\
0 & =u^2-2 \times 9.8 \times 4.9 \\
\Rightarrow \quad u^2 & =\sqrt{(9.8)^2} \Rightarrow u=9.8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\begin{aligned}
\text { Impulse } & =m[v-(-u)] \\
& =1[9.8 \sqrt{2}+9.8]=9.8(\sqrt{2}+1) \\
& =9.8 \times 2.4=23.52 \mathrm{~N}\mathrm{-s}
\end{aligned}
\)
\(
\text { Average force }=\frac{\text { Impulse }}{\text { Time }}=\frac{23.52}{0.1}=235.2 \mathrm{~N}
\)

Hint

(a) Here, mass, \(m=5 \mathrm{~kg}\)

Change in velocity, \(\Delta \mathbf{v}=v_f-v_i=[(10-2) \hat{\mathbf{i}}+(6-6) \hat{\mathbf{j}}]=8 \hat{\mathbf{i}}\)
Change in momentum \(=m \Delta \mathbf{v}=5[8 \hat{\mathbf{i}}]=40 \hat{\mathbf{i}} \mathrm{kg}-\mathrm{ms}^{-1}\)

Hint

(b) Impulse is defined as rate of change of momentum. For change in momentum to be minimum,
\(
\begin{aligned}
\frac{d}{d t}\left(20 t^2-40 t\right) & =0 \\
40 t-40 & =0 \Rightarrow t=1 \mathrm{~s}
\end{aligned}
\)

Hint

\(
\text { (a) } F_{\mathrm{av}}=m a_{\mathrm{av}}=m \frac{\Delta v}{\Delta t}=\frac{0.5[2-(-2)]}{10^{-3}}=2000 \mathrm{~N}
\)

Hint

(c) From law of conservation of momentum, \(p_i=p_f\)
and initial momentum, \(p_i=m u=m(0)=0\)
\(\therefore p_f\) should also be zero.
Hence, other piece will move in negative \(x\)-direction.

Hint

(d) From the law of conservation of momentum, total initial momentum \(=\) total final momentum
\(
\begin{aligned}
& \Rightarrow \quad m_1 \mu_1+m_2 u_2=m v_1+m v_2 \\
& \therefore \quad 0.1 \times 0+50 \times 0=0.1 \times 100+50\left(-v_2\right) \\
& \Rightarrow \quad 0=10-50 v_2 \\
& \therefore \quad v_2=\frac{10}{50}=0.2 \mathrm{~ms}^{-1} \\
&
\end{aligned}
\)

Hint

(b) From the given figure,

Initial velocity, \(\mathbf{v}_i=\left(-20 \sin 30^{\circ} \hat{\mathbf{i}}-20 \cos 30^{\circ} \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}\)
Final velocity, \(\mathbf{v}_f=\left(-20 \sin 30^{\circ} \hat{\mathbf{i}}+20 \cos 30^{\circ} \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}\)
Change in velocity, \(\Delta \mathbf{v}=\mathbf{v}_f-\mathbf{v}_i\)
[/latex]
\begin{aligned}
& \Delta \mathbf{v}=\left(-20 \sin 30^{\circ} \hat{\mathbf{i}}+20 \cos 30^{\circ} \hat{\mathbf{j}}\right) \\
& \Delta \mathbf{v}=2 \times 20 \cos 30^{\circ} \hat{\mathbf{j}} \quad-\left(-20 \sin 30^{\circ} \hat{\mathbf{i}}-20 \cos 30^{\circ} \hat{\mathbf{j}}\right) \\
& \Delta \mathbf{v}=2 \times 20 \times \frac{\sqrt{3}}{2} \hat{\mathbf{j}} \Rightarrow \Delta \mathbf{v}=20 \sqrt{3} \mathbf{j}
\end{aligned}
[/latex]

Magnitude, \(|\Delta \mathbf{v}|=20 \sqrt{3} \mathrm{~m} / \mathrm{s}\)

Hint

Horizontal component of \(\mathbf{F}\) is

\(
\begin{aligned}
F_H & =F \cos 45^{\circ} \\
& =(8)\left(\frac{1}{\sqrt{2}}\right)=4 \sqrt{2} \mathrm{~N}
\end{aligned}
\)
and vertical component of \(\mathbf{F}\) is
\(
\begin{aligned}
F_V & =F \sin 45^{\circ} \\
& =(8)\left(\frac{1}{\sqrt{2}}\right)=4 \sqrt{2} \mathrm{~N}
\end{aligned}
\)

Hint

Component perpendicular to the plane,

\(
w_{\perp}=w \cos 30^{\circ}=(10) \frac{\sqrt{3}}{2}=5 \sqrt{3} \mathrm{~N}
\)
and component parallel to the plane,
\(
w_{||}=w \sin 30^{\circ}=(10)\left(\frac{1}{2}\right)=5 \mathrm{~N}
\)

Hint

The object is in equilibrium.
\(
\begin{aligned}
& \Sigma F_x=0 \\
& \therefore 8+4 \cos 60^{\circ}-F_2 \cos 30^{\circ}=0 \\
& 8+2-F_2 \frac{\sqrt{3}}{2}=0 \\
& F_2=\frac{20}{\sqrt{3}} \mathrm{~N} \\
& \text { Similarly, } \quad \Sigma F_y=0 \\
& \therefore F_1+4 \sin 60^{\circ}-F_2 \sin 30^{\circ}=0 \\
& F_1+\frac{4 \sqrt{3}}{2}-\frac{F_2}{2}=0 \\
& F_1=\frac{F_2}{2}-2 \sqrt{3} \\
& =\frac{10}{\sqrt{3}}-2 \sqrt{3} \\
& F_1=\frac{4}{\sqrt{3}} \mathrm{~N} \\
&
\end{aligned}
\)

Hint

Resolving the tension \(T_1\) along horizontal and vertical directions. As the body is in equilibrium,
\(\mathrm{T}_1 \sin 60^{\circ}=4 \times 9.8 \mathrm{~N} \dots(i)\) and
\(T_1 \cos 60^{\circ}=T_2 \dots(ii)\)
From Eq. (i), we get
\(
T_1=\frac{4 \times 9.8}{\sin 60^{\circ}}=\frac{4 \times 9.8 \times 2}{\sqrt{3}}=45.26 \mathrm{~N}
\)
Putting this value in Eq. (ii), we get
\(
T_2=T_1 \cos 60^{\circ}=45.26 \times 0.5=22.63 \mathrm{~N}
\)

Note Tension is a type of force produced in strings.

Hint

\(A C=0.5 \mathrm{~m}, B C=0.3 \mathrm{~m}\)
\(\therefore A B=0.4 \mathrm{~m}\)
and if \(\angle B A C=\theta\)
Then, \(\cos \theta=\frac{A B}{A C}=\frac{0.4}{0.5}=\frac{4}{5}\)
and \(\quad \sin \theta=\frac{B C}{A C}=\frac{0.3}{0.5}=\frac{3}{5}\)
Here, the object is in equilibrium under three concurrent forces. So, we can apply Lami’s theorem,
\(
\frac{F}{\sin \left(180^{\circ}-\theta\right)}=\frac{8}{\sin \left(90^{\circ}+\theta\right)}=\frac{T}{\sin 90^{\circ}}
\)

\(
\begin{aligned}
\frac{F}{\sin \theta} & =\frac{8}{\cos \theta}=T \\
T & =\frac{8}{\cos \theta} \\
& =\frac{8}{4 / 5}=10 \mathrm{~N} \\
F & =\frac{8 \sin \theta}{\cos \theta} \\
& =\frac{(8)(3 / 5)}{(4 / 5)}=6 \mathrm{~N}
\end{aligned}
\)

Hint

Since, the block is permanently at rest, it is in equilibrium. Net force on it should be zero. In this case, only two forces are acting on the block.
(i) Weight \(=m g\) (downwards)
(ii) Contact force (resultant of normal reaction and friction force) applied by the wedge on the block.

For the block to be in equilibrium, these two forces should be equal and opposite. Therefore, force exerted by the wedge on the block is \(m g\) (upwards).

Explanation:

From Newton’s third law of motion, force exerted by the block on the wedge is also \(\mathrm{mg}\) but downwards. The result can also be obtained in a different manner. The normal force on the block is \(R\) or \(N=m g \cos \theta\) and the friction force on the block is \(f=m g \sin \theta(\operatorname{not} \mu m g \cos \theta)\) because it is not the case of limiting friction.
These two forces are mutually perpendicular.
\(\therefore\) Net contact force would be \(\sqrt{N^2+f^2}\) or \(\sqrt{(m g \cos \theta)^2+(m g \sin \theta)^2}\) which is equal to \(m g\).

Hint

The equilibrium condition is obtained when the net force acting on the body is zero. All of the above choice is true.

Hint

For horizontal equilibrium: \(T_3 \sin 60^{\circ}=T_2 \sin 30^{\circ} \therefore T_3=\frac{T_2}{\sqrt{3}}\)
Now, for vertical equilibrium:
\(
T_2 \cos 30^{\circ}+T_3 \cos 60^{\circ}=T_1 \dots(1)
\)
\(
\mathrm{T}_1=\mathrm{mg}=10 \times 10=100 \mathrm{~N}
\)
Substituting the value of \(T_1\) and \(T_3\) in equation (1), we get
\(
\begin{aligned}
\frac{\sqrt{3} T_2}{2}+\frac{T_2}{\sqrt{3}} \times \frac{1}{2} & =100 \\
\frac{8 T_2}{4 \sqrt{3}} & =100 \\
T_2 & =50 \sqrt{3} \mathrm{~N}
\end{aligned}
\)

Hint

Considering the forces acting in the vertically upward direction, \(2 \mathrm{~T} \cos 60^{\circ}=\mathrm{W}\)
\(
\mathrm{W}_{\max }=\mathrm{T}_{\max }=20 \mathrm{~N}
\)
In the horizontal direction, the component of the forces cancel out.

Hint

Since both blocks (left and right, \(m\)) are similar and tension in the strings are \(\mathrm{mg}\). It has zero acceleration, as the system is in equilibrium.
For the block on the left is \(T-m g=m a=m(0)\)
For the blockin the right is \(T-m g=m a=m(0)\)
Equilibrium of \(\sqrt{2} m: 2 T \cos \theta=\sqrt{2} m g\)
\(
\begin{aligned}
& \Rightarrow 2 m g \cos \theta=\sqrt{2} m g \\
& \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\
& \Rightarrow \theta=45^{\circ}
\end{aligned}
\)

Hint

\(
\begin{aligned}
T_1 \cos 60^{\circ}= & T_2 \cos 30^{\circ} \\
\frac{T_1}{2} & =\frac{\sqrt{3} T_2}{2}
\end{aligned}
\)‘
\(
T_1=\sqrt{3} T_2 \dots(i)
\)
\(
\begin{aligned}
& \therefore \quad \frac{\sqrt{3} T_1}{2} \times A G=\frac{T_2}{2} \times B G \\
& \Rightarrow \quad \frac{A G}{B G}=\frac{T_2}{\sqrt{3} T_1} \\
& \text { But } \quad \frac{T_2}{T_1}=\frac{1}{\sqrt{3}} \quad \text { [from Eq. (i)] }\\
& \text { Hence, } \quad \frac{A G}{B G}=\frac{1}{3} \\
&
\end{aligned}
\)

Hint

For equilibrium of body,
\(
m g=2 T \cos \theta
\)
\(
T=\frac{m g}{2 \cos \theta}
\)

For the string to be horizontal,
\(
\begin{aligned}
\theta & =90^{\circ} \\
T & =\frac{m g}{2 \cos 90^{\circ}} \\
\Rightarrow \quad T & =\infty
\end{aligned}
\)

Hint

As shown the figure below

\(
\begin{aligned}
T_1 \cos 30^{\circ} & =T_2 \cos 30^{\circ} \\
\text { Let, }T_1 & =T_2=T
\end{aligned}
\)
Again, \(T_1 \sin 30^{\circ}+T_2 \sin 30^{\circ}=10 \Rightarrow 2 T \sin 30^{\circ}=10\)
\(\Rightarrow \quad 2 T \cdot \frac{1}{2}=10 \Rightarrow T=10 \mathrm{~N}\)
Thus, the tension in section \(B C\) and \(B F\) are \(10 \mathrm{~N}\) and \(10 \mathrm{~N}\), respectively.

Hint

The given figure can be drawn as

Taking component of forces,
\(
\begin{aligned}
R \cos \theta=M g \Rightarrow & R \cos 60^{\circ}=M g \dots(i) \\
\text { and } & R \sin 60^{\circ}=T \dots(ii)
\end{aligned}
\)
and
\(
R \sin 60^{\circ}=T
\)

By Eqs. (i) and (ii), we get
\(
\begin{aligned}
\Rightarrow & & \tan 60^{\circ} & =\frac{T}{M g} \Rightarrow T=M g \tan 60^{\circ} \\
\text { or } & & T & =60 \times g \times \sqrt{3}=103.9 \mathrm{kgf}
\end{aligned}
\)

Hint

(a) The mass of the frame is \(m=30 \mathrm{~kg}\)
The mass of the man is \(M=50 \mathrm{~kg}\)
The man is standing on the frame. So, the total mass of the system is \(M+m=50 \mathrm{~kg}+30 \mathrm{~kg}=80 \mathrm{~kg}\)
Now the frame is on a pulley, the other end of the rope is on the hand of the man. The man is pulling the rope to make the system to be in equilibrium.
Let the man is pulling the rope with a force \(\mathrm{F}\) and the tension on the rope will be \(\mathrm{T}\). Again, the frame is also attached to one end of the rope. The weight of the frame is downward. So, we have another tension \(T\) on the rope which is also acting upward.
For the system to be in equilibrium, the downward force on the system should be equal to the upward force.
Total downward force \(m\) the system will be,
\(
\begin{array}{l}
(M+m) g = 80 g
\end{array}
\)
The total upward force will be,
\(
T+T=2 T
\)
Now, these two forces will be equal.
\(
\begin{array}{l}
2 T=80 g \\
T=\frac{80 g}{2} \\
T=40 g
\end{array}
\)
Since the force with which the man is pulling the rope is equal to the tension on the rope, we can say that the system will be in equilibrium when the man pulls the rope with a force \(40 g\).

Hint

(d) Let \(x\) be distance from \(A\) to \(O\) and \(L\) be the total length of string. Then, ratio of tensions is
\(
\frac{T_1}{T_2}=\frac{x}{L}=\frac{1}{3}
\)

Hint

The resultant force will be
\(
F_R=\left|\mathbf{F}_R\right|=\sqrt{3^2+4^2+2 \times 3 \times 4 \cos 90^{\circ}}=5 \mathrm{~N}
\)

Hint

(a) Let equal forces \(F_1=F_2=F\) newton
Angle between the forces, \(\theta=60^{\circ}\)
Resultant force, \(R=40 \sqrt{3} \mathrm{~N}\)
Now,
\(
R=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos \theta}
\)
\(
\begin{array}{l}
\therefore \quad 40 \sqrt{3}=\sqrt{F^2+F^2+2 F F \cos 60^{\circ}} \\
\Rightarrow \quad 40 \sqrt{3}=\sqrt{2 F^2+2 F^2 \times \frac{1}{2}} \\
\Rightarrow \quad F=40 \mathrm{~N} \\
\end{array}
\)

Hint

Let us find the net magnitude of forces \(F_1\) and \(F_2\) which are perpendicular to each other and this net force be written as \(F_{12}\) so,
\(
F_{12}=\sqrt{F^2+F^2}
\)
\(F_{12}=\sqrt{2} F\) Which will be in direction just opposite to that of \(F_3\).
Now, two forces acting on the body of mass \(m\) respectively \(F_{12}=\sqrt{2} F\) and \(F_3=F\).
Both forces are in opposite direction,
Hence net force act on the body of mass \(m\) is,
\(
\begin{array}{l}
F_{\text {net }}=\sqrt{2} F-F \\
\Rightarrow F_{\text {net }}=(\sqrt{2}-1) F
\end{array}
\)
Now, let us assume that net acceleration is denoted by \(a^{\prime}\) then by newton’ second law we have:
\(
(\sqrt{2}-1) F=m a^{\prime}
\)
We also know that, \(F=m a\) put this value in above equation, we get,
\(
(\sqrt{2}-1) a=a^{\prime}
\)
Net acceleration is \(a^{\prime}=(\sqrt{2}-1) a\)

Hint

Let the additional force \(\mathrm{F}\) be directed along the positive \({x}\)-direction Taking \({x}\)-components, the total should be zero
\(
x+1 \cos 60^{\circ}+2 \cos 60^{\circ}-4 \sin 30^{\circ}=0
\)
\(
x=0.5 \mathrm{~N}
\)

Hint

(c) Apply Lami’s theorem at \(O\),
\(
\begin{aligned}
\frac{T_1}{\sin 150^{\circ}} & =\frac{T_2}{\sin 120^{\circ}}=\frac{10}{\sin 90^{\circ}}=\frac{10}{1}=10 \\
\therefore \quad T_1 & =10 \sin 150^{\circ}=10 \times \frac{1}{2}=5 \mathrm{~N} \\
T_2 & =10 \sin 120^{\circ} \\
& =10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} \mathrm{~N}
\end{aligned}
\)