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Chapter 4: Problem 60

A crate of potatoes of mass \(18.0 \mathrm{kg}\) is on a ramp with angle of incline \(30^{\circ}\) to the horizontal. The coefficients of friction are \(\mu_{\mathrm{s}}=0.75\) and \(\mu_{\mathrm{k}}=0.40 .\) Find the frictional force (magnitude and direction) on the crate if the crate is sliding up the ramp.

Short Answer

Expert verified
Answer: The magnitude of the frictional force on the crate is 61.12 N, and the direction is downwards along the ramp.

Step by step solution

01

Calculate gravitational force

First, we need to find the gravitational force acting on the crate. The gravitational force (weight) can be calculated using the formula: \(F_g = m \times g\) where \(F_g\) is the gravitational force, \(m\) is the mass of the crate, and \(g\) is the acceleration due to gravity (approximately \(9.81 \, m/s^2\)). Given mass \(m = 18.0 \, kg\), we have: \(F_g = 18.0 \, kg \times 9.81 \, m/s^2 = 176.58 \, N\)
02

Calculate normal force

Now that we have the gravitational force on the crate, we can calculate the normal force acting on the crate. The normal force is perpendicular to the ramp surface and can be calculated using the formula: \(F_n = F_g \times \cos{\theta}\) where \(F_n\) is the normal force and \(\theta\) is the angle of inclination. Given angle \(\theta = 30^{\circ}\), we have: \(F_n = 176.58 \, N \times \cos{30^{\circ}} = 152.81 \, N\)
03

Calculate frictional force

To find the frictional force acting on the crate when it is sliding up the ramp, we can use the formula for kinetic friction: \(F_f = \mu_k \times F_n\) where \(F_f\) is the frictional force and \(\mu_k\) is the coefficient of kinetic friction. Given coefficient of kinetic friction \(\mu_k = 0.40\), we have: \(F_f = 0.40 \times 152.81 \, N = 61.12 \, N\) Since the crate is sliding up the ramp, the frictional force acts in the opposite direction, i.e., down the ramp. Therefore, the magnitude of the frictional force on the crate is \(61.12 \, N\), and the direction is downwards along the ramp.

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Most popular questions from this chapter

A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?
An \(85-\mathrm{kg}\) skier is sliding down a ski slope at a constant velocity. The slope makes an angle of \(11^{\circ}\) above the horizontal direction. (a) Ignoring any air resistance, what is the force of kinetic friction acting on the skier? (b) What is the coefficient of kinetic friction between the skis and the snow?
Oliver has a mass of \(76.2 \mathrm{kg} .\) He is riding in an elevator that has a downward acceleration of \(1.37 \mathrm{m} / \mathrm{s}^{2} .\) With what magnitude force does the elevator floor push upward on Oliver?
While an elevator of mass 832 kg moves downward, the tension in the supporting cable is a constant \(7730 \mathrm{N}\) Between \(t=0\) and \(t=4.00 \mathrm{s},\) the elevator's displacement is \(5.00 \mathrm{m}\) downward. What is the elevator's speed at \(t=4.00 \mathrm{s} ?\)
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of \(560 \mathrm{N}\) at an angle of \(15^{\circ}\) with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge.
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