Fh = force up of hands at x = xh from axle
Fa = force up from wheel axle at x = 0
Wb = 67 at xb from axle
Wc = 511 at xc from axle
then
Fa+Fh = 67+511
Fh * xh = 67 * xb + 511 * xc
The wheels, axle, and handles of a wheelbarrow weigh W = 67 N. The load chamber and its contents weigh WL = 511 N. The drawing shows these two forces in two different wheelbarrow designs. To support the wheelbarrow in equilibrium, the man’s hands apply a force to the handles that is directed vertically upward. Consider a rotational axis at the point where the tire contacts the ground, directed perpendicular to the plane of the paper. Find the magnitude of the man’s force for both designs.
1 answer
1 answer