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  • A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force (in N) on B to make bo

A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force (in N) on B to make both A and B move together, is

 

Option: 1

36


Option: 2

24


Option: 3

12


Option: 4

48


Answers (1)

best_answer

As we learned

 

Motion of Two Bodies one Resting on the Other(2) -

 

  • Case 2:- A force F is applied to the lower body,

Let's discuss possible 4 situations under this case:-

          

2.    If friction is present between A and B only and

            F' (Pseudo force on body A) < Fl (limiting friction between body A and B):-

                             Both the body will move together with common acceleration,

                               which is given by  a=\frac{F}{m+M}

                And the value of F' will be 

                 F'=ma=\frac{F}{m+M}=\frac{mF}{m+M} \\ and \ F_l=\mu _smg

                        as \ F' < F_l \Rightarrow \frac{mF}{m+M} <\mu _smg\\ \Rightarrow F < \mu _s(m+M))g

                        So both bodies will move together with acceleration, 

                             a_A=a_B=\frac{F}{m+M} \ if \ \ F < \mu _s(m+M))g

3.     If friction is present between A and B only and F > F'l(limiting friction between body A and B):-

               F'_l= \mu _s(m+M))g

 Both the body will move with different acceleration. Here force of kinetic friction μkmg will oppose the motion of B while will cause the motion of A.

 

ma_A=\mu _kmg\\ i.e \ \ a_A= u _kg

similarly

\begin{array}{c}{F-F_{k}=M a_{B}} \\ {\text { i.e. } \quad a_{B}=\frac{\left[F-\mu_{k} m g\right]}{M}}\end{array}

As both, the bodies are moving in the same direction,

Acceleration of body A relative to B will be given by:-

a=a_{A}-a_{B}=-\left[\frac{F-\mu_{k} g(m+M)}{M}\right]

              Negative sign implies that relative to B, A will move backward and will fall it after time,

                   t=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 M L}{F-\mu_{k} g(m+M)}}

 

So, on the basis of this we can solve this,

Maximum friction i.e. limiting friction between A and B, is F_{l} = 12 N.

If F is the maximum value of force applied on lower body such that both body move together, it means pseudo force on upper body is just equal to limiting friction

F = F_{l} \Rightarrow m \left ( \frac{F}{m+M} \right )=12\Rightarrow \left ( \frac{4}{4+8} \right )F=12\Rightarrow F=36 N

Posted by

shivangi.shekhar

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