Aperson pushing a horizontal, uniformly loaded, 25.30 kg wheelbarrow of length l is attempting to get it over a step of height h=0.430r, 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. px=

The minimum force is 357.12 N.

Explanation:

Given that,

Load = 25.30 kg

Height h = 0.430 R

We need to calculate the height

We need to calculate the value of x

Using formula of x

We need to calculate the horizontal force

Using balance equation of torque

Put the value into the formula

Hence, The minimum force is 357.12 N.

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

overcharge is the right answer system