Aperson pushing a horizontal, uniformly loaded, 23.90 kg23.90 kg wheelbarrow of length is attempting to get it over a step of height ℎ=0.=0.370r, where is the wheel's radius. the center of gravity of the wheelbarrow is in the center of the wheelbarrow. what is the horizontal component pxpx of the minimum force p⃗ p→ necessary to push the wheelbarrow over the step? the gravitational acceleration is =9.81 m/s2.

The horizontal component of the minimum force is 144.24 N.

Explanation:

Given that,

Loaded = 23.90 kg

Height = 0.370 R

Where, R=wheel's radius

We need to calculate the acceleration

Using formula of acceleration

Put the value into the formula

We need to calculate the horizontal component of the minimum force

Using moment about center of point of contact

Hence, The horizontal component of the minimum force is 144.24 N.

answer; how far out of place which the object finished relative to the starting location;