Question

A person 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=

Answer 1

Answer:

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

$h=(R-0.430 R)$

$h=(1-0.430)R$

$h=0.57R$

We need to calculate the value of x

Using formula of x

$x=\sqrt{R^2-(0.57R)}$

$x=0.821R$

We need to calculate the horizontal force

Using balance equation of torque

$mgx=F\times h$

$F=\dfrac{mgx}{h}$

Put the value into the formula

$F=\dfrac{25.30\times9.8\times0.821 R}{0.57 R}$

$F=357.12\ N$

Hence, The minimum force is 357.12 N.