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Chapter 4: Problem 58

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 at rest.

Short Answer

Expert verified
Answer: The frictional force acting on the crate is 114.68 N, directed up the ramp.

Step by step solution

01

Calculate the gravitational force

First, we need to find the gravitational force acting on the crate. This can be calculated by the formula: \(F_g = m \times g\) where \(m = 18.0\,\text{kg}\) is the mass of the crate and \(g = 9.8\,\text{m/s}^2\) is the acceleration due to gravity. \(F_g = 18.0\,\text{kg} \times 9.8\,\text{m/s}^2 = 176.4\,\text{N}\)
02

Resolve the gravitational force into components

Now, we need to resolve the gravitational force into two components: one parallel to the ramp (\(F_{g\parallel}\)) and the other perpendicular to the ramp (\(F_{g\perp}\)). To do this, we will use the geometry of the ramp and the angle given: \(F_{g\parallel} = F_g \cdot \sin(\theta)\) \(F_{g\perp} = F_g \cdot \cos(\theta)\) where \(\theta = 30^\circ\) is the angle of incline. \(F_{g\parallel} = 176.4\,\text{N} \cdot \sin(30^\circ) = 88.2\,\text{N}\) \(F_{g\perp} = 176.4\,\text{N} \cdot \cos(30^\circ) = 152.91\,\text{N}\)
03

Calculate the normal force

The normal force (\(F_n\)) acting on the crate is equal to the component of the gravitational force that acts perpendicular to the ramp: \(F_n = F_{g\perp} = 152.91\,\text{N}\)
04

Calculate the maximum static frictional force

Since the crate is at rest, we will use the coefficient of static friction (\(\mu_s = 0.75\)) to find the maximum frictional force (\(F_f\)) acting on the crate. This can be calculated by the formula: \(F_f = \mu_s \times F_n\) \(F_f = 0.75 \times 152.91\,\text{N} = 114.68\,\text{N}\)
05

Determine the direction of the frictional force

As the crate is at rest, the frictional force must balance the component of the gravitational force that acts parallel to the ramp. Thus, the frictional force (-\(F_f\)) and the \(F_{g\parallel}\) must have opposite directions. Therefore, the frictional force is 114.68 N acting up the ramp, counteracting the component of the gravitational force acting down the ramp.

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Most popular questions from this chapter

A cord, with a spring balance to measure forces attached midway along, is hanging from a hook attached to the ceiling. A mass of \(10 \mathrm{kg}\) is hanging from the lower end of the cord. The spring balance indicates a reading of \(98 \mathrm{N}\) for the force. Then two people hold the opposite ends of the same cord and pull against each other horizontally until the balance in the middle again reads \(98 \mathrm{N} .\) With what force must each person pull to attain this result?
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In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of \(2.0 \mathrm{kg} .\) Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is \(0.25 .\) (a) If she pushes the puck with a force of \(5.0 \mathrm{N}\) in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes \(7.5 \mathrm{N} .\) Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of \(6.0 \mathrm{N}\) on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of \(6.0 \mathrm{N}\) forward. Is the acceleration of the puck during contact more, the same, or less than on Earth? Explain. (tutorial: rough table)
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