Question: A block of mass $2 \mathrm{Kg}$ placed on a long frictionless horizontal table is pulled horizontally by a constant force $\mathbf{F}$. It is found to move $10 \mathrm{~m}$ in the first two seconds. Find the magnitude of the force.

Answer: $\text { Mass, } m=2 \mathrm{kg}$,$\text { distance } s=10 \mathrm{~m}$,$\text { initial velocity } u=0$

Time taken $t=2 \mathrm{sec}$

Then using, $s=u t+1 / 2 a t^2$

u = 0, $s=1 / 2 a t^2$

Putting the values of $s$ and $t$

$a=5 \mathrm{~m} / \sec ^2$

Then using, $a =F / m \text { or } F=m a$

$F =2 \times 5 =10 \mathrm{~N}$

$s=0+1 / 2 a t^2 \quad(\text { for } u=0)$

$t=\sqrt{\frac{25}{a}}-\text { (4) }$

putting the value of $t$ in (3)

$v=\sqrt{\frac{25 F}{m}}$

putting the values of $S, F,$ and $m$,

$v =\sqrt{\left(\frac{2 \times 3 \times 7 \times 10^5}{2 \times 10^7}\right)} =0.45 \mathrm{m} / \mathrm{sec}$

Question: A block of mass $m$ slides along a floor while a force of magnitude $\mathbf{F}$ is applied to it at an angle as shown in figure. The coefficient of kinetic friction is $\mu$. Then, what will be the value of acceleration of the block.

Answer: $N=m g-F \sin \theta \text {-(1) }$

For moving body.

$F=\cos \theta-f=m a \text {-(2) }$

Where $f=$ frictional farce

$f=\mu N=\mu(m g-F \sin \theta)\text {-(3) }$

With (2) and (3)

$m a=F \cos \theta-\mu(m g-F \sin \theta)\text {-(4) }$

Or, $a=F / m \cos \theta-\mu(g-F / m \sin \theta)\text {-(5) }$

Answer:

For block at rest,

$F=m g \sin \theta$

$F=\text { frictional force }$

Frictional force $=m g \sin \theta$

$=4 \times 9.8 \times \left(\frac{1}{2}\right)$

$=19.6 \mathrm{M}$

Question: The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, what should be the value of angle $\theta$.

Answer For system in equilibrium,

$T_1=m g ; \quad T_2=m g -(1)$

$T_1 \cos \theta_1+T_2 \cos \theta=\sqrt{2} m g -(2)$

$T_1 \sin \theta_1=T_2 \sin \theta \quad \text { -(3) }$

From (1) and (3)

$m g \sin \theta_1=m g \sin \theta$

$\text { or } \quad \theta_1=\theta -(4)$

Using (4) and (1). (2) can be written as, $m g \cos \theta+m g \cos \theta=\sqrt{2} m g$

or $2 \cos \theta=\sqrt{2}$

$\theta=45^{\circ}$

Question: What is the maximum value of the force F such that the block shown in the arrangement, does not move?

Answer:

$N=m g+F \sin 60^{\circ} -(1)$

Frictional force $=F \cos 60^{\circ} -(2)$

Frictional force $=\mu N$

$=\mu\left[m g+F \sin 60^{\circ}\right] -(3)$

From (3) and (1)

$\mu\left[m g+F \sin 60^{\circ}=F \cos 60^{\circ}\right].$

Or $F =\frac{\mu m g}{\cos 60^{\circ}-\mu \sin 60^{\circ}}$

$=\frac{(1 / 2 \sqrt{3}) \sqrt{3} \times 9.8}{(1 / 2)-{(1 / 2 \sqrt{3}) \times(\sqrt{3} / 2)}}$

$=\frac{4.9}{\left(1/2-1 / 4\right)}$ = 19.6 N

Net force at. $P$ must be equal to zero for this,

$F=R=2 T \sin \theta-(1)$

For mass $m$, with no vertical movement

$T \sin \theta=m g -(2)$

$T \cos \theta=m a -(3)$

from (1) and (3)

$m a =\frac{F}{2} \cot \theta$

$a=\frac{F}{2 m}\left(\frac{x}{\sqrt{q^2-x^2}}\right)$

Answer:

(a) Moving down

$F+m g \sin 3 i=f$ (frictional force) $-(1)$

$f=\mu N \quad$ ', $\mu=$ coefficient of friction $-(2)$

$N=m g \cos 30^{\circ} \quad-(3)$

From (2) and (3)

$f=\mu\left(m g \cos 30^{\circ}\right)-(4)$

From (1) and (4)

$F=m g\left(\mu \cos 30^{\circ}-\sin 30^{\circ}\right)$

Putting the values of $m, g$ and $\mu$

$F=2 \times 9.8\left[(\sqrt{3} / \sqrt{2}) \times \cos 30^{\circ}-\sin 30^{\circ}\right]$

= 10.99 N

Answer:

(b) Moving up.

$F =m g \sin 30^{\circ}+f$

$=m g \sin 3{ }^{\circ}+\mu m g \cos 30^{\circ}$

$=m g\left(\sin 30^{\circ}+\mu \cos 30^{\circ}\right)$

Putting the values of $m, g$, and $\mu$

$F=2 \times 9.8\left[(1 / 2)+(\sqrt{3} / \sqrt{2})\left(\frac{\sqrt{3}}{2}\right)\right]$

= 30.58 N

Question: In the figure, the blocks A, B and C have masses 3kg, 4 kg and 8 kg respectively. The coefficient of sliding friction between any two surfaces is 1.25. A is held at rest by a massless rigid rod fixed to the wall, while B and C are connected bya alight flexible cord passing around a fixed frictionless pulley, Find the force F necessary to drag C along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the figure, i.e, B on C and A on B, is maintained throughout

Masses of the bodies

$A=m_A=3 \mathrm{kg}$, $B=m_B=4 \mathrm{kg}$, and $C=m_C=8 \mathrm{kg}$

Coefficient of kinetic friction $\mu=0.25$ (for any two surfaces)

Friction fore between $A$ and $B$

$f_{A B}=\mu m_A g -(1)$

Friction force between $B$ and $C$

$f_{B C}=\mu\left(m_A+m_B\right) g \text {-(2) }$

Friction force between $C$ and surface

$f_{c s}=\mu\left(m_A+m_B+m_C\right) g -(3)$

Answer :

If we pull C with Force

F then foces will act on B and C will be like this,

T=$f_{AB} + f_{BC} -(4)$

F =$T + f_{BC} + f_{CS} -(5)$

F =$f_{AB} + 2 f_{BC} + f_{CS} -(6)$

Putting the values of $f_{AB}, f_{BC}, f_{CS}$

from (1), (2) and (3)

$F=\mu\left(4 m_A+3 m_B+m_C\right) g$

Putting the values of $m_A, m_B$, and $m_C$

$F=0.25(4 \times 3+3 \times 4+8) 9.8$

= 78.4 N

Solution: For equilibrium,

N = mg cos$\alpha$ -(1)

And frictional force, f = $\mu N$ -(2)

Also, f = mg sin$\alpha$ -(3)

From (1),(2) and (3)

$\mu (mg \cos\alpha) = mg \sin\alpha$

$\mu = tan\alpha$

For $\mu = \frac{1}{3}$

tan$\alpha=\frac{1}{3}$ or $cot \alpha = 3$