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Chapter 6: Problem 89

A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68,\) and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) \(222 \mathrm{~N}\), (b) \(334 \mathrm{~N},\) (c) \(445 \mathrm{~N},\) and (d) \(556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

Short Answer

Expert verified
Frictional forces: (a) 222 N, (b) 334 N, (c) 311.36 N, (d) 311.36 N. Cabinet moves in (c) and (d).

Step by step solution

01

Determine the Maximum Static Friction Force

The maximum static friction force can be calculated using the formula: \( f_s = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. Here, \( N = 556 \, \text{N} \). So, \( f_s = 0.68 \cdot 556 = 378.08 \, \text{N} \). This means any force less than or equal to 378.08 N will not move the cabinet.
02

Evaluate Attempt (a)

For the force of 222 N, since 222 N is less than 378.08 N, the frictional force will equal the applied force. Thus, the frictional force is 222 N.
03

Evaluate Attempt (b)

For the force of 334 N, since 334 N is also less than 378.08 N, the frictional force will equal the applied force. So, the frictional force is 334 N.
04

Evaluate Attempt (c)

For the force of 445 N, since 445 N exceeds 378.08 N, static friction is overcome and the cabinet starts moving. Thus, the frictional force transitions to kinetic friction: \( f_k = \mu_k \cdot N = 0.56 \cdot 556 = 311.36 \, \text{N} \). Hence, the frictional force is 311.36 N.
05

Evaluate Attempt (d)

For the force of 556 N, since 556 N also exceeds 378.08 N, the cabinet is moved with kinetic friction. The frictional force is the kinetic friction force: 311.36 N.
06

Identify Moving Attempts

Attempts (c) and (d) result in motion because they exceed the maximum static friction (378.08 N). Therefore, these attempts move the cabinet.

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Key Concepts

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

maximum static friction
Static friction keeps objects stationary, even when a small force is applied. It must be overcome to start motion. The maximum static friction force is the highest force that static friction can exert before moving begins. This force is calculated using the formula:\[ f_s = \mu_s \cdot N \]where:
  • \( f_s \) is the maximum static friction force,
  • \( \mu_s \) is the coefficient of static friction,
  • \( N \) is the normal force.
In the given exercise, the normal force is the weight of the cabinet, which is 556 N. With a static friction coefficient of 0.68, the maximum static friction force is:\[ f_s = 0.68 \times 556 = 378.08 \, \text{N} \]This means any applied force less than or equal to 378.08 N will not move the cabinet.
kinetic friction force
Once the static friction is overcome, the object begins to move. At this point, kinetic friction comes into play. The kinetic friction force is usually less than static friction, making it easier to keep the object moving once it's started. The kinetic friction force can be determined using:\[ f_k = \mu_k \cdot N \]where:
  • \( f_k \) is the kinetic friction force,
  • \( \mu_k \) is the coefficient of kinetic friction,
  • \( N \) is the normal force.
In this scenario, with a kinetic friction coefficient of 0.56, once movement starts, the kinetic friction force is:\[ f_k = 0.56 \times 556 = 311.36 \, \text{N} \]This force maintains as long as the cabinet is in motion.
coefficient of static friction
The coefficient of static friction, \( \mu_s \), is a dimensionless value that represents the frictional force between two stationary surfaces. It quantifies how much force is necessary to start moving an object at rest. Typically, the larger the coefficient, the more force is needed to initiate motion. In our example, the coefficient is 0.68, which indicates a relatively higher level of resistance from the static friction between the cabinet and the floor. This value is essential for calculating the threshold at which the object begins to slide.
normal force
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In many situations, like the one in this exercise, the normal force equals the weight of the object when it's resting on a horizontal surface without additional vertical forces. Here, the filing cabinet's weight is 556 N, and accordingly, the normal force is also 556 N. Understanding the normal force is critical, as it directly influences both static and kinetic friction calculations. This relationship ensures that the frictional forces are proportional to the normal force acting on the object.

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Most popular questions from this chapter

What is the terminal speed of a \(6.00 \mathrm{~kg}\) spherical ball that has a radius of \(3.00 \mathrm{~cm}\) and a drag coefficient of \(1.60 ?\) The density of the air through which the ball falls is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\).

A locomotive accelerates a 25-car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v,\) where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is \(30 \mathrm{~km} / \mathrm{h},\) the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2} .\) (a) \(\mathrm{What}\) is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h} ?\)

A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg} .\) Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the net force on the bicycle from the road.

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle \(200 \mathrm{~m}\) slope in the Alps, a skier reduced his top-to-bottom time from 61 s with standard skis to 42 s with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a \(3.0^{\circ}\) slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

Calculate the ratio of the drag force on a jet flying at \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a propdriven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km} .\) Assume that the airplanes have the same effective cross-sectional area and drag coefficient \(C\).
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