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Chapter 4: Problem 39

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{渭}_ ext{s} = 0.760 ext{}) and the normal force.\[ F_{ ext{static}} = ext{渭}_ ext{s} imes F_n = 0.760 imes 588 = 447.68 ext{ N} \]
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{渭}_ ext{k} = 0.410 ext{}) and the normal force.\[ F_{ ext{kinetic}} = ext{渭}_ ext{k} imes F_n = 0.410 imes 588 = 241.08 ext{ N} \]
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 bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is \(80.0 \mathrm{~kg},\) and the hill is inclined at \(15.0^{\circ}\) with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

The earth exerts a gravitational force on a falling raindrop, pulling it down. Does the raindrop exert a gravitational force on the earth, pulling it up? If so, is this force greater than, less than, or equal to the force that the earth exerts on the raindrop? Provide a reason for your answer. Problem A raindrop has a mass of \(5.2 \times 10^{-7} \mathrm{~kg}\), and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.

A block is pressed against a vertical wall by a force \(\vec{P}\), as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity, the magnitude of the force being different in the two cases, while the directional angle \(\theta\) is the same. Kinetic friction exists between the block and the wall. (a) Is the block in equilibrium in each case? Explain. (b) In each case what is the direction of the kinetic frictional force that acts on the block? Why? (c) In each case is the magnitude of the frictional force the same or different? Justify your answer. (d) In which case is the magnitude of the force \(\overrightarrow{\mathrm{P}}\) greater? Provide a reason for your answer. The weight of the block is \(39.0 \mathrm{~N}\), and the coefficient of kinetic friction between the block and the wall is \(0.250\). The direction of the force \(\overrightarrow{\mathrm{P}}\) is \(\theta=30.0^{\circ}\). Determine the magnitude of \(\overrightarrow{\mathrm{P}}\) when the block slides up the wall and when it slides down the wall. Check to see that your answers are consistent with your answers to the Concept Questions.

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?

Two forces \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{F}} \mathrm{B}\) are applied to an object whose mass is \(8.0 \mathrm{~kg}\). The larger force is \(\vec{F}_{A} .\) When both forces point due east, the object's acceleration has a magnitude of \(0.50 \mathrm{~m} / \mathrm{s}^{2} .\) However, when \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}\) points due east and \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) points due west, the acceleration is \(0.40 \mathrm{~m} / \mathrm{s}^{2}\), due east. Find (a) the magnitude of \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}\) and (b) the magnitude of \(\vec{F}_{B}\)
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