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A \(60.0\) -kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are \(0.760\) and \(0.410\), respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

Short Answer

Expert verified
447.68 N to start moving; 241.08 N to slide at constant speed.

Step by step solution

01

Understand the Problem

We need to determine two forces: (a) the force required to just start moving the crate (overcome static friction) and (b) the force required to slide the crate at a constant speed (overcome kinetic friction). The crate weighs 60.0 kg and the coefficients of static and kinetic friction are given.
02

Calculate the Normal Force

The normal force (F_n) is the force exerted by the floor, which balances the gravitational force. Since the floor is level, the normal force is equal to the weight of the crate. Calculate this using the formula: F_n = m imes g, where m = 60.0 ext{ kg} and g = 9.8 ext{ m/s}^2.\[ F_n = 60.0 \times 9.8 = 588 ext{ N} \]
03

Calculate Static Friction Force

To find the force needed to start the crate moving, calculate the maximum static friction using the coefficient of static friction (egin{equation*} ext{μ}_ ext{s} = 0.760 ext{}) and the normal force.\[ F_{ ext{static}} = ext{μ}_ ext{s} imes F_n = 0.760 imes 588 = 447.68 ext{ N} \]
04

Calculate Kinetic Friction Force

To find the force needed to keep the crate moving at a constant speed, calculate the kinetic friction force using the coefficient of kinetic friction (egin{equation*} ext{μ}_ ext{k} = 0.410 ext{}) and the normal force.\[ F_{ ext{kinetic}} = ext{μ}_ ext{k} imes F_n = 0.410 imes 588 = 241.08 ext{ N} \]
05

Interpret Results

Therefore, the force needed to just start the crate moving is approximately 447.68 N, and the force needed to slide it at constant speed is approximately 241.08 N.

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Key Concepts

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

Static Friction
Static friction is the force that resists the initial movement of an object resting on a surface. This force must be overcome to set an object into motion. Its magnitude depends on two things: the coefficient of static friction and the normal force. In the exercise, we deal with a crate that needs an initial push. This push must be strong enough to overcome static friction.
The coefficient of static friction is a measure of how much grip the surfaces have on each other. For the crate example, this value is 0.760. Since the normal force is 588 N, the static friction can be calculated as follows:
  • Maximum static friction = Coefficient of static friction x Normal force
  • \(F_{\text{static}} = 0.760 \times 588 = 447.68\, \text{N} \)
Therefore, a force of approximately 447.68 N is required to just start pushing the crate.
Kinetic Friction
Once an object begins moving, the force resisting its continued motion is called kinetic friction. Unlike static friction, kinetic friction usually remains constant regardless of speed. This means, once the object starts sliding, you only need to overcome kinetic friction to keep it moving at a constant speed. The coefficient for kinetic friction in our example is 0.410.
To calculate the kinetic friction force:
  • Kinetic friction = Coefficient of kinetic friction x Normal force
  • \(F_{\text{kinetic}} = 0.410 \times 588 = 241.08\, \text{N} \)
Thus, only around 241.08 N of force is needed to maintain the crate's movement across the dock once it has started moving.
Normal Force
The normal force is an important concept as it acts perpendicular to the surface an object rests on. It is vital in friction calculations because both static and kinetic friction forces are derived from it. In the example problem, the normal force is the gravitational force that acts on the crate from the dock floor. With our crate weighing 60.0 kg, we can determine the normal force as follows:
  • Normal force = Mass x Acceleration due to gravity
  • \( F_{n} = 60.0 \times 9.8 = 588\, \text{N} \)
Knowing the normal force allows us to calculate both static and kinetic frictional forces required for moving the crate.
Force Calculation
In physics, calculating forces is essential in understanding motion and resistance. In this case, we look at two major forces: the force needed to overcome static friction to initiate movement, and the force to overcome kinetic friction to sustain movement.
The steps for calculating the required force to start and continue moving the crate are as follows:
  • Determine the normal force, which we've found is 588 N.
  • Calculate the static friction force: 447.68 N.
  • Calculate the kinetic friction force: 241.08 N.
Thus, the initial push of around 447.68 N is needed to move the crate from rest, and only 241.08 N is necessary to maintain its motion at a steady pace. Understanding these forces helps in planning the effort needed to manipulate objects on various surfaces.

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Most popular questions from this chapter

A car is driving up a hill. (a) Is the magnitude of the normal force exerted on the car equal to the magnitude of its weight? Why or why not? (b) If the car drives up a steeper hill, does the normal force increase, decrease, or remain the same? Justify your answer. (c) Does the magnitude of the normal force depend on whether the car is traveling up the hill or down the hill? Give your reasoning. Problem A car is traveling up a hill that is inclined at an angle of \(\theta\) above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) \(\theta=15^{\circ}\) and \((\mathrm{b}) \theta=35^{\circ} .\) Check to see that your answers are consistent with your answers to the Concept Questions.

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of \(+11 \mathrm{~m} / \mathrm{s}\) in a time of \(28 \mathrm{~s}\). The combined mass of the boat and trailer is 410 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car?

The central ideas in this problem are reviewed in Multiple-Concept Example \(9 .\) One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches \(47.0 \mathrm{~N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

Refer to Concept Simulation \(4.4\) at for background relating to this problem. The drawing shows a large cube (mass \(=25 \mathrm{~kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\vec{P}\). A small cube \((\mathrm{mass}=4.0 \mathrm{~kg})\) is in contact with the front surface of the large cube and will slide downward unless \(\vec{P}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71\). What is the smallest magnitude that \(\overrightarrow{\mathrm{P}}\) can have in order to keep the small cube from sliding downward?

A 350 -kg sailboat has an acceleration of \(0.62 \mathrm{~m} / \mathrm{s}^{2}\) at an angle of \(64^{\circ}\) north of east. Finc the magnitude and direction of the net force that acts on the sailboat.

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