A man pushes a cylinder of mass m1 with the help of a plank of mass m2. There is no slipping at any contact. The horizontal component of the force applied by the man is F.Finda)the accelerations of the plank and the centre of mass of the cylinder,andb)the magnitudes and directions of frictional force at contact points.

Dear Bharath Devasani,

Ans:- Now draw the FBD of two bodies.

The cylinder in the bottom will have the following set of forces 

f1=force of friction between the two bodies and is towards right for the cylinder and towards left for the plank. 

f2=force of friction between the ground and the cylinder and is towards left.

F=applied horizontal force on the plank

a1= Acceleration of the plank towards the right

a= Acc of the cylinder towards right

ß= Angular Acc of the cylinder

Then the eqns  of motion are

F-f1=M2 a1.........................(1)

f1 - f2=M1 a........................(2)

(f1+f2)r=Iβ...........................(3) (where I= momont of inertia= 1/2 M2 r²)

And the conditions of non slleeping are

a1=a+ βr..................(4)(At the top cotact of two bodies)

a2=βr..............(5)(At the contact of the cylinder and the ground)

Solving these Eqsns we get

a=4F/(8M2+3M1)

a1=8F/(8M2+3M1)

Please feel free to post as many doubts on our discussion 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 Bharat Devasani !!!

Regards,

Askiitians Experts
Soumyajit Das IIT Kharagpur