Question

A person pushing a horizontal, uniformly loaded, 23.90 kg23.90 kg wheelbarrow of length ????L is attempting to get it over a step of height ℎ=0.370????,h=0.370R, 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 PxPx of the minimum force P⃗ P→ necessary to push the wheelbarrow over the step? The gravitational acceleration is ????=9.81 m/s2.

Answer 1

Answer:

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

$R^2=a^2+(R-h)^2$

$a^2=R^2-(R-h)^2$

Put the value into the formula

$a^2=R^2-(R-0.370R)^2$

$a=\sqrt{0.631R^2}$

$a=0.776R$

We need to calculate the horizontal component of the minimum force

Using moment about center of point of contact

$P_{x}(R-h)=\dfrac{mg}{2}\times a$

$P_{x}(R-0.370)=\dfrac{23.90\times9.8}{2}\times0.776R$

$P_{x}=\dfrac{23.90\times9.8\times0.776R}{2(R-0.370)}$

$P_{x}=144.24\ N$

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