Question

A roller of mass 300kg and of radius 50cm lying on horizontal floor is rest
and of radius 50cm lying on horizontal floor is resting against a step of height
20cm . The minimum horizontal force to be applied on the roller passing through its centre to turn the
roller on to the step is​

Answer 1

Answer:

● Net torque about the line joining the step size and cyclinder = 0 Hence,

F x (50 – 20) = Mg x R

We get,

F = 5Mg/3 => 5000N

would be the toppling horizontal force required to turn the roller onto the step.

Answer 2

Answer:

5000 N

Explanation:

since it would under go toppling

mg× R = F × (50-30)