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Q.29

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Found in: Page 155

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# A $4000$ kg truck is parked on a $15°$ slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is $0.90$.

The friction force on the truck is $9,140.45N$.

See the step by step solution

## Step 1: Content Introduction

The friction force is equal to the product of the coefficient of friction and the normal force. The force that stops one solid object from sliding or rolling over another is known as friction. Frictional forces, such as the traction needed to walk without slipping, can be beneficial, but they can cause a lot of resistance to movement.

## Step 2: Content Explanation

Let the horizontal weight force is ${W}_{y}=W\mathrm{cos}\left(\theta \right)$ and vertical weight force is ${W}_{x}=W\mathrm{sin}\left(\theta \right)$.

According to Newton's second law of motion, the friction force can be written as ${f}_{r}=\mu N$

Here we are given

localid="1648206782774" $\mu =0.9\phantom{\rule{0ex}{0ex}}m=4000kg\phantom{\rule{0ex}{0ex}}g=9.8m/{s}^{2}$

Substitute the vales , here, ${f}_{r}$ is the friction force, $\mu$ is the coefficient of static between friction force, $N$ is the normal force

${f}_{r}=\mu N\phantom{\rule{0ex}{0ex}}{f}_{r}=\mu .W\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}{f}_{r}=\mu .mg\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}{f}_{r}=0.9×4000×9.8×\mathrm{sin}15°\phantom{\rule{0ex}{0ex}}{f}_{r}=9,140.45N$