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In Figure, the coefficient of friction between the floor and the body B is 0.1. The coefficient of friction between bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is $\frac{m}{2}$ m, and of B, it is m. Which of the following statements is true?

(a) The bodies will move together if F = 0.25 mg.
(b) The body A will slip concerning B if F = 0.5 mg.
(c) The bodies will move together if F = 0.5 mg.
(d) The bodies will be at rest if F = 0.1 mg.
(e) The maximum value of F for which the two bodies will move together is 0.45 mg.

Answers (1)

The correct answer is: -

(a) The bodies will move together at if $\overrightarrow{ F} = 0.25 mg$ .

(b) The body A will slip concerning B if $\overrightarrow{ F} = 0.5 mg$ .

(d)The bodies will be at rest if $\overrightarrow{ F} = 0.1 mg$ .

(e) The maximum value $\overrightarrow{ F}$ for which the two bodies will move together is 0.45 mg.

Explanation: By opt (e) the max. The force by which bodies move together is 0.45 mg, Newton.

mA = $\frac{m}{2}$ = mB = m

Consider the acceleration of the bodies A and B to be ‘a.

Both the bodies A & B will keep moving by force F until the force of friction between their surface is larger or equal to 0 than the force acting on A.

$a=\frac{F-f_{1}}{m_{A}-m_{B}}=\frac{F-f_{1}}{\frac{m}{2}+m}=\frac{2(F-f_{1})}{3m}$

Thus, the force on

$A=m_{A}a=\frac{m}{2}\frac{\left ( F-f_{1} \right )}{3m}$

Hence, the force on A is

$F_{AB}=\frac{\left ( F-f_{1} \right )}{3}$

Body A will move along with body B only if F_AB is equal to or smaller than f₂.

Hence, $F_{AB}=f_{2}$

$\mu N =\frac{\left ( F-f_{1} \right )}{3}$

$0.2\times m_{A}g =\frac{\left ( F-f_{1} \right )}{3}$ ……… (i)

N is the reaction force by B on A

$f_{1} = \mu NB = \mu (m_{A} + m_{B}) g$ ……… (here, N_B is the normal reaction on B along with A by the surface)

$0.1 \times (m_{A} + m_{B}) g$

$f_{1} = 0.1 \times \frac{3}{2} mg = 0.15mg$ ………… (ii)

Now,

$F - f_{1} = 3 \times 0.2 mAg$ ……….. [from (i)]

$F - 0.15mg = 0.6 \times \frac{m}{2} g$

$F_{max} = 0.3mg + 0.15mg = 0.45mg$ …………. (iii)

Thus, the max force on B is 0.45mg, therefore, A & B can move together.

Hence, opt (e).

Both the bodies A & B will keep moving by force F until the force of friction between their surface is larger or equal to 0 than the force acting on A, i.e., 0.45 mg Newton, hereby opt (c) is rejected.

Now, for opt (d), the minimum force which can move A & B together is,

$F_{in} \geq f_{1}+f_{2}$

$\geq 0.15 mg+0.2\times \frac{m}{2}g$ ….. [from (i) & (ii)]

$F_{in}\geq 0.25 mg$ Newton

Since 0.1mg<0.25 mg, opt (d) is verified which states that the body will be at rest if $\overrightarrow{F} = 0.1 mg.$

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