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Grade 11General Physics

Q. A solid cylinder of mass 6kg lies on a rough horizontal table.The coefficient of friction between the cylinder and the surface is 0.2 . A constant force acts horizontally on the cylinder.The line of action of force F is at a height 2R/3 above the centre of the cylinder.find the max value of F if cylinder rolls without slipping?

Ans.108 N.

Please help.

Profile image of Raj Deol
15 Years agoGrade 11
Answers icon

5 Answers

Profile image of jay gohel
15 Years ago

here the max force F will be there when friction just starts acting .

hence, considering the free body diag we have,   F-f=ma ( f is friction ). ___1

here friction will act opposite to dir of motion.    

now considering newtons 2nd law of rotation we have F x 2R/3-f x r =I x (angular accn.)______2

I=mr2 /2 and due to rolling angular caan = a/R ____3

solving 1,2,&3

we have F= 9 µmg

substitute values and get F = 108 N

Profile image of Abhishek Sharma
12 Years ago

hey i am solving like this but my answer is not matching .

can you please elabrote your answer

Profile image of KAPIL MANDAL
11 Years ago
Thank you sir for this answer. Thank you very much.
Profile image of Animesh Bhardwaj
9 Years ago
It should be F+ff=ma,........,..........................answer content should not be less than 100 characters...... So I m doing it
Profile image of Kushan Choudhury
6 Years ago
Static Friction will be max. For max. Force =>f=umg
Torque eqn:
F×2R/3 + f ×R =I × alpha where i=mr^2/2
Alpha =acm/r for rolling
Also F-f=ma
Or,F-umg =ma
Solving these eqn |F| =9umg =108N

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