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Q.29

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 155

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Short Answer

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 9,140.45 N.

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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 Wy=W cosθ and vertical weight force is Wx=W sinθ.

According to Newton's second law of motion, the friction force can be written as fr=μN

Here we are given

localid="1648206782774" μ=0.9 m=4000 kgg=9.8 m/s2

Substitute the vales , here, fr is the friction force, μ is the coefficient of static between friction force, N is the normal force

fr=μNfr=μ . W sinθfr=μ. mg sinθfr=0.9×4000×9.8×sin 15°fr=9,140.45 N

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