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Chapter 6: Problem 18

\(A\) stubborn, \(120 \mathrm{kg}\) mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of \(800 \mathrm{N}\). The coefficients of friction between the mule and the ground are \(\mu_{1}=0.8\) and \(\mu_{1}=0.5 .\) Is the farmer able to move the mule?

Short Answer

Expert verified
No, the farmer is not able to move the mule as the force he applies (800 N) is less than the maximum static friction (940.8 N).

Step by step solution

01

Calculate normal force

First, apply the formula for normal force, which is equal to weight for this scenario. Using the formula Normal Force = Mass x Gravity, Substitute Mass = 120 kg and Gravity = 9.8 m/s^2 to get Normal Force = 120 kg x 9.8 m/s^2 = 1176 N.
02

Calculate the maximum static friction

Next, calculate the maximum static friction using the formula Maximum Static Friction = Coefficient of static friction x Normal Force. Substitute Coefficient of static friction = 0.8 and Normal Force = 1176 N to get Maximum Static Friction = 0.8 x 1176 N = 940.8 N.
03

Compare forces

Now, compare the force applied by the farmer with the maximum static friction calculated in step 2. If the force applied by the farmer (800 N) is greater than the maximum static friction (940.8 N), then the farmer will be able to move the mule. However, in this case, it's less than the static friction.

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Key Concepts

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

Normal Force
In physics, the normal force plays a crucial role when understanding how objects interact with surfaces. It's the force exerted by a surface to support the weight of an object resting on it. Imagine if you place a book on a table; the table applies an upward force, preventing the book from falling through.
This force is often equal to the object's weight acting perpendicular to the surface. For the mule, we calculate the normal force using the formula:
  • Normal Force = Mass 脳 Gravity
  • where Mass = 120 kg and Gravity = 9.8 m/s虏
So, Normal Force = 120 kg 脳 9.8 m/s虏 = 1176 N. This force helps us understand how tightly the mule is pressed against the ground, influencing the friction between the mule and the ground.
Coefficient of Friction
The coefficient of friction is a value that represents the frictional force between two objects. It has no units and can vary depending on the surfaces involved. Typically, there are two coefficients to consider:
  • Static coefficient (\( \mu_{s} \)): Represents the frictional force necessary to start moving an object at rest.
  • Kinetic coefficient (\( \mu_{k} \)): Represents the friction for moving objects.
In our problem:
  • \( \mu_{s} = 0.8 \) (static coefficient) is the value used to determine if the farmer can start moving the mule, because it is initially at rest.
  • The static friction formula is: Maximum Static Friction = \( \mu_{s} \) x Normal Force.
Thus, Maximum Static Friction = 0.8 脳 1176 N = 940.8 N.
Force Comparison
After understanding the forces at play, we evaluate if the farmer can move the mule. We must compare the maximum static friction, which we've calculated to be 940.8 N, with the force the farmer applies, which is 800 N.
To determine the motion:
  • If the applied force is greater than the maximum static friction, the object would start moving.
  • If the applied force is less, the object remains stationary.
In this scenario, the farmer's force of 800 N falls short of overcoming the maximum static friction of 940.8 N, which means that the mule will not budge. The force comparison is essential when predicting movement.
Newton's Laws of Motion
Newton's Laws of Motion provide the fundamental principles needed to understand the behavior of objects in motion.
  • First Law (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion unless acted upon by a net external force.
  • Second Law: The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass (\( F = ma \)).
  • Third Law: For every action, there is an equal and opposite reaction.
In the case of the stubborn mule, Newton's First Law is especially relevant. The mule remains at rest because the applied force is not sufficient to overcome the static friction keeping it stationary. To set the mule in motion, the farmer's force would need to surpass the maximum static friction, creating a net force resulting in acceleration, according to Newton's Second Law.

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Most popular questions from this chapter

A \(1.0 \mathrm{kg}\) ball hangs from the ceiling of a truck by a 1.0 -m-long string. The back of the truck, where you are riding with the ball, has no windows and has been completely soundproofed. The truck travels along an exceedingly smooth test track, and you feel no bumps or bounces as it moves. Your only instruments are a meter stick, a protractor, and a stopwatch. a. The driver tells you, over a loudspeaker, that the truck is either at rest, or it is moving forward at a steady speed of \(5 \mathrm{m} / \mathrm{s} .\) Can you determine which it is? If so, how? If not, why not? b. Next, the driver tells you that the truck is either moving forward with a steady speed of \(5 \mathrm{m} / \mathrm{s}\), or it is accelerating at \(5 \mathrm{m} / \mathrm{s}^{2} .\) Can you determine which it is? If so, how? If not, why not? c. Suppose the truck has been accelerating forward at \(5 \mathrm{m} / \mathrm{s}^{2}\) long enough for the ball to achieve a steady position. Does the ball have an acceleration? If so, what are the magnitude and direction of the ball's acceleration? d. Draw a free-body diagram that shows all forces acting on the ball as the truck accelerates. e. Suppose the ball makes a \(10^{\circ}\) angle with the vertical. If possible, determine the truck's velocity. If possible, determine the truck's acceleration.

You are given the dynamics equations that are used to solve a problem. For each of these, you are to a. Write a realistic problem for which these are the correct equations. b. Draw the free-body diagram and the pictorial representation for your problem. c. Finish the solution of the problem. $$\begin{array}{l} T-0.20 n-(20 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) \sin 20^{\circ} \\\=(20 \mathrm{kg})\left(2.0 \mathrm{m} / \mathrm{s}^{2}\right) \\\n-(20 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) \cos 20^{\circ}=0\end{array}$$

An artist friend of yours needs help hanging a 500 lb sculpture from the ceiling. For artistic reasons, she wants to use just two ropes. One will be \(30^{\circ}\) from vertical, the other \(60^{\circ} .\) She needs you to determine the smallest diameter rope that can safely support this expensive piece of art. On a visit to the hardware store you find that rope is sold in increments of \(\frac{1}{8}\) -inch diameter and that the safety rating is 4000 pounds per square inch of cross section. What diameter rope should you buy?

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about \(1 \mathrm{~m}\) as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield. These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about \(5 \mathrm{~mm}\) a. \(A 60 \mathrm{~kg}\) person is in a head-on collision. The car's speed at impact is \(15 \mathrm{~m} / \mathrm{s}\). Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys. b. Estimate the net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag. c. How do these two forces compare to the person's weight?

A \(1500 \mathrm{kg}\) car skids to a halt on a wet road where \(\mu_{\mathrm{x}}=0.50\) How fast was the car traveling if it leaves \(65-\) m-long skid marks?
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