Question

In the figure, what magnitude of force Upper F Overscript right-arrow EndScripts applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height h = 0.396 m? The wheel's radius is r = 0.503 m and its mass is m = 1.25 kg.

Answer 1

here,

The component of the force F perpendicular to the moment arm = Fsin(theta).

The torque that the force applies is therefore = F r sin(theta)

In order for the force to lift the wheel, it must at provide at least as much torque as the force of gravity.

The component of gravity perpendicular to the moment arm = mgcos(theta), so the torque it applies = mgrcos(theta).

Frsin(theta) = mgrcos(theta)

F = mgcot(theta)

Now you have to do some geometry to find (theta):

sin(theta) = (r-h)/r = 1 - h/r, so (theta) = arcsin(1 - h/r)

F = mgcot( arcsin(1 - h/r) )

F = (1.25)(9.8) cot( arcsin( 1 - 0.396/0.503) )

F = 56.3 N

the magntude of force is 56.3 N