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```        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.
7 years ago

jay gohel
4 Points
```										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
```
7 years ago
Abhishek Sharma
18 Points
```										hey i am solving like this but my answer is not matching .
```
4 years ago
KAPIL MANDAL
132 Points
```										Thank you sir for this answer. Thank you very much.
```
2 years ago
Animesh Bhardwaj
24 Points
```										It should be F+ff=ma,........,..........................answer content should not be less than 100  characters...... So I m doing it
```
one year ago
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