Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • In Figure, the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is m/2 and of B is m.

In Figure, the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is \frac{m}{2} and of B is m. Which of the following statements are true?
 
(a) The bodies will move together if F = 0.25 mg.
(b) The body A will slip with respect to 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 with respect to B if \overrightarrow{ F} = 0.5 mg.

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

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

Explanation: By opt (e) the max. 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}

The body A will move along with body B only if FAB is equal 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, NB is the normal reactn 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.

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question