Question

A person pushing a horizontal, uniformly loaded, 26.40 kg wheelbarrow of length L is attempting to get it over a step of height h = 0.470R, where R is the wheel's radius. The center of gravity of the wheelbarrow is in the center of the wheelbarrow. 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. PE

Answer 1

Sol:

Given
M=26.4 kg
h=0.47 R

sinΘ = 1 - 0.47
= 0.53
Θ =sin-¹( 0.53)
Θ= 32.005°

then the Minimum horizontal force will be

Px = 26.40 kg * 9.81m/s² / tan32.005º
= 414.381 N
=414.4 N

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