Question

A person pushing a horizontal, uniformly loaded, 21.95 kg wheelbarrow of length L is attempting to get it over a step of height h=0.510R, where R is the wheel's radius. The center of gravity of the wheelbarrow is in the center of the wheelbarrow.

1 OL

What is the horizontal component Px of the minimum force P→ necessary to push the wheelbarrow over the step? The gravitational acceleration is g=9.81 m/s2.

Px=

Answer 1

The free-body diagram for the wheel is as follows, m=21.95 kg h=(0.510R) The remaining hieht of the center of the gravity of the wheel from the height of the step from the above diagram is, h = R-h =R-(0.510R) = R(1-0.51) = 0.49 R The distance moved by the wheel of the center of the gravity of the wheel is, x=VR- = R - (0.49R)* = R -(0.2401) R2 =R? (1-0.2401) = VR (0.7599) =(0.8717)R Apply the torque on the wheel as follows, The torque to the gravitational force (counter clockwise) is equal to the torque due to the horizontal component force (clockwise), that is, mg = P. 2 Here, P. is the horizontal component of the force P. Rearrange the above equation for P, and substitute the values in the equation. 12h (21.95 kg 9.81 m/s (0.8717) R 2(0.510R) = 184.02 N Therefore, the horizontal component of the force P, is 184.02 N.