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Strike441 [17]
3 years ago
7

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=

Physics
1 answer:
Vlad [161]3 years ago
7 0

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.

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Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

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