# Answer to Question #3220 in Mechanics | Relativity for ZherVin

Question #3220

Find the force F which when applied at an horizontal angle of 15 degrees to the center of a wheel of radius R and weight W will pull it over a step of height y <r

Expert's answer

We assume that force applied at an horizontal angle of 15 degrees up, otherwise it wouldn't pull the wheel over a step, it would pull it down. This is a torque problem, with the pivot point being at the corner of the step that the wheel will rotate around to get on top of the step. The applied force must provide a torque greater than the torque provided by the weight.

So the torque provided by force

edge of the step

F_min = W/cos(15)

So the torque provided by force

**F: F*cos(15)*r > mg*r**- the torque provided by the weight when when center of the wheel is at distance r from the verticaledge of the step

**F > mg/cos(15), mg = W**

F > W/cos(15)F > W/cos(15)

F_min = W/cos(15)

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