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Chapter 5: Problem 33

You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp (Fig. E5.33). Both boxes move together at a constant speed of \(15.0 \mathrm{~cm} / \mathrm{s}\). The coefficient of kinetic friction between the ramp and the lower box is 0.444 , and the coefficient of static friction between the two boxes is 0.800 . (a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

Short Answer

Expert verified
The force which the person needs to exert to move the boxes at constant speed is the sum of the frictional forces on the lower and upper boxes. The magnitude of the frictional force on the upper box is equal to the static frictional force between the boxes and its direction is downwards along the ramp.

Step by step solution

01

Find the total weight of the system

The weight of the system is the sum of the weights of both boxes. If \(m_1\) and \(m_2\) are the masses of the two boxes, the total weight \(W_{total}\) is given by the equation \(W_{total} = m_1 g + m_2 g\) where \(g\) is the acceleration due to gravity, which is approximately \(9.81 \mathrm{m/s^2}\). In this problem, we don't know the masses of the boxes, but that won't prevent us from solving the problem.
02

Find the frictional force between the ramp and the lower box

The frictional force \(F_{f1}\) between the ramp and the lower box is given by the equation \(F_{f1} = \mu_{k1} m_2 g\), where \(\mu_{k1}\) is the coefficient of kinetic friction between the ramp and the lower box, and \(m_2\) is the mass of the lower box. This frictional force opposes the movement of the boxes and is directed upward along the ramp.
03

Find the frictional force between the two boxes

The frictional force \(F_{f2}\) between the two boxes is given by the equation \(F_{f2} = \mu_{s} m_1 g\), where \(\mu_{s}\) is the static frictional coefficient between the boxes, and \(m_1\) is the mass of the upper box. This frictional force tends to prevent the top box from slipping off the lower box, and is directed downward along the ramp.
04

Find the total force exerted by the person

Since the boxes are moving with a constant speed, we can set up the following equation of motion: \(F_{person} - F_{f1} - F_{f2} = 0\) where \(F_{person}\) is the force exerted by the person, \(F_{f1}\) is the kinetic frictional force on the lower box, and \(F_{f2}\) is the static frictional force on the upper box. Solving for \(F_{person}\), we find \(F_{person} = F_{f1} + F_{f2}\) which is the total force exerted by the person.
05

Find the magnitude and direction of friction on the upper box

The magnitude of the friction on the upper box is just \(F_{f2}\) and this force is directed downwards along the ramp. We already calculated \(F_{f2}\) in Step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictional Force
Frictional force is the force that opposes the motion of an object in contact with a surface. It's a crucial concept in physics, particularly when analyzing situations involving motion-like objects on a ramp. Friction arises due to the interaction between the surface and the object. This frictional force depends on two key factors:
  • The nature of the surfaces in contact: Rough surfaces tend to have higher friction.
  • The normal force: The force exerted by the surface perpendicular to the object.
Frictional force can be helpful, such as when it prevents slipping between a person's shoes and the ground, or it can oppose motion, like when it resists the effort to slide a heavy box across the floor.
It's important to understand that friction can occur in two main types: kinetic friction when an object is moving, and static friction when it's at rest but there is an impending motion. Each type has its unique coefficient which helps quantify the amount of friction present.
Kinetic Friction
Kinetic friction is the force that opposes the movement of two surfaces sliding past each other. If you picture a box sliding down a ramp, the kinetic friction acts in the opposite direction to the box's movement. The amount of kinetic friction is determined by two factors:
  • The coefficient of kinetic friction (\(\mu_k\)): This is a dimensionless value representing the "stickiness" between surfaces.
  • The normal force (\(N\)): The force pressing the two surfaces together. This typically equals the weight component perpendicular to the surface.
The kinetic friction force (\(F_k\)) can be calculated using the equation: \[F_k = \mu_k \cdot N\]This force reduces the speed of moving objects and is a key consideration in situations like the one in the exercise, where two boxes slide down a ramp.
Understanding kinetic friction is essential for predicting how objects move when they're already in motion and how much force is required to maintain that movement at a constant speed.
Static Friction
Static friction comes into play when an object is at rest, but there is an impending force trying to move it. It's the reason why objects don鈥檛 start moving until a certain force threshold is crossed, like when you're trying to push a heavy piece of furniture.Static friction is usually larger than kinetic friction, making it more challenging to start an object moving from a rest position. Its magnitude can adjust to match the applied force, up to a maximum value defined by:
  • The coefficient of static friction (\(\mu_s\)): Often higher than the coefficient for kinetic friction.
  • The normal force (\(N\)): Like kinetic friction, it's the component of the weight acting perpendicular to the surface.
The static friction force (\(F_s\)) has a maximum calculated by:\[F_s \leq \mu_s \cdot N\]In this exercise, static friction acts between the two boxes, preventing the top box from slipping off the bottom one.
Recognizing the role of static friction helps in understanding how forces must overcome this friction to cause movement, which is critical for applications in many real-world scenarios.

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Most popular questions from this chapter

Some sliding rocks approach the base of a hill with a speed of \(12 \mathrm{~m} / \mathrm{s} .\) The hill rises at \(36^{\circ}\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and \(0.65,\) respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

An \(8.00 \mathrm{~kg}\) block of ice, released from rest at the top of a 1.50-m-long friction less ramp, slides downhill, reaching a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of \(10.0 \mathrm{~N}\) parallel to the surface of the ramp?

A steel ball with mass \(m\) is suspended from the ceiling at the bottom end of a light, 15.0 -m-long rope. The ball swings back and forth like a pendulum. When the ball is at its lowest point and the rope is vertical, the tension in the rope is three times the weight of the ball, so \(T=3 m g .\) (a) What is the speed of the ball as it swings through this point? (b) What is the speed of the ball if \(T=m g\) at this point, where the rope is vertical?

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

If the coefficient of static friction between a table and a uniform, massive rope is \(\mu_{\mathrm{s}},\) what fraction of the rope can hang over the edge of the table without the rope sliding?
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