Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
abhishek kumar sharma Grade: 12

Q: As shown in the fig. a small block of mass 'm' is placed on a bigger block of mass 'M'. Friction coefficient between the blocks is mu and the surface between bigger block and surface is frictionless. Find the maximum force which can be applied on smaller box in the horizontal forward direction to prevent the relative motion between the blocks.

8 years ago

Answers : (2)

Pratham Ashish
17 Points


if both boxes move together their acceleration,

   a = F / (m+M)

 for big box , it will move only due to friction force

 it moves with max accel. when the friction will be max.

    a M =  µ. mg  / M                                                        ( max friction =   µ. mg  )

 if both boxes move together ,

 a =  a M

 F / (m+M) = µ. mg  / M     

 F =  (m+M) µ. mg  / M     

 it will be the max force for both the boxes move together


8 years ago
Ramesh V
70 Points

forces on block M and m are

on balancing forces we get,

N2 =mg

N = N2 +Mg



so from above

F = (m+M)a

putting above gives : umg = Ma

so   F=u(M+M)mg / M

where u is coeff. of friction btn teo blocks


Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and

we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.


Naga Ramesh

IIT Kgp - 2005 batch

8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details