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Chapter 3: Problem 66

A 20 -kg wagon is pulled along the level ground by a rope inclined at \(30^{\circ}\) above the horizontal. A friction force of \(30 \mathrm{~N}\) opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and \((b)\) an acceleration of \(0.40 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
(a) 34.64 N, (b) 43.85 N

Step by step solution

01

Analyze the Forces for Constant Speed

To maintain constant speed, the pulling force must counteract the friction force. The pulling force can be represented as \( F \), which has a horizontal component \( F \cos(\theta) \) and a vertical component \( F \sin(\theta) \). The vertical component doesn't affect horizontal motion on a level ground. Thus, for constant speed, \( F \cos(30^{\circ}) = 30 \mathrm{~N} \) because it needs to exactly balance the horizontal friction force.
02

Solve for Pulling Force for Constant Speed

Solve for \( F \) using the equation from Step 1:\[ F \cos(30^{\circ}) = 30 \mathrm{~N} \]\[ F = \frac{30 \mathrm{~N}}{\cos(30^{\circ})} \]\[ \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \], so:\[ F = \frac{30}{\frac{\sqrt{3}}{2}} \approx \frac{60}{\sqrt{3}} \approx 34.64 \mathrm{~N} \].
03

Consider Additional Force for Acceleration

For part (b), the wagon not only has to overcome the friction but also needs to accelerate. Therefore, the net force \( F_{net} \) must also account for the force needed to accelerate (\( ma \)), where \( m \) is the mass and \( a \) is the acceleration.\[ F_{net} = F \cos(30^{\circ}) - 30 = 20 \times 0.4 \]
04

Solve for Pulling Force with Acceleration

Calculate the net force using \( ma = 20 \times 0.4 = 8 \mathrm{~N} \) and use it to determine \( F \):\[ F \cos(30^{\circ}) - 30 = 8 \]\[ F \cos(30^{\circ}) = 38 \]Substitute \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \):\[ F = \frac{38 \mathrm{~N}}{\cos(30^{\circ})} = \frac{38}{\frac{\sqrt{3}}{2}} \approx \frac{76}{\sqrt{3}} \approx 43.85 \mathrm{~N} \].

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Key Concepts

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

Forces and Motion
Understanding forces and motion is crucial to solving physics problems like the wagon exercise. Forces are essentially pushes or pulls acting on an object, and they determine how that object moves. When it comes to forces involved in motion on Earth, they include gravity, friction, tension, and applied forces. In this case, the wagon is influenced by an applied force through the rope and an opposing friction force.

Motion occurs as a direct result of these forces interacting with objects. If the forces are balanced, the object will either remain at rest or move with constant velocity. Conversely, if the forces are unbalanced, the object will accelerate. This concept is central to answering the original exercise, where the wagon is moving at constant speed or accelerating due to different force interactions.
  • A force is needed to start moving or stop a wagon.
  • Friction opposes the direction of motion.
  • The applied force must be greater than the opposing force to cause acceleration.
Kinematics
Kinematics is the branch of physics that describes the motion of objects. It focuses on parameters like velocity, acceleration, and displacement without necessarily considering the causes of motion, which are addressed by dynamics. In the problem, you observe the wagon moving with constant speed or accelerating, illuminating key concepts of kinematics.

Constant speed implies that the velocity of an object does not change over time. There's no acceleration involved, and net force is zero here. Meanwhile, acceleration refers to the rate of change of velocity. In the problem when the wagon accelerates, a net force must act on it alongside overcoming friction.
  • Displacement is the change in the position of the object.
  • Velocity is the speed with a direction.
  • Acceleration is the change in velocity over time.
Recognizing these concepts helps decode how different forces influence the outcomes in kinematics equations and real-world problems.
Newton鈥檚 Laws of Motion
Newton's Laws of Motion form the foundation for explaining the nature of forces and motion. The exercise about the wagon utilizes primarily the first and second laws. The first law (Law of Inertia) indicates that an object at rest stays at rest, and an object in motion stays in motion at a constant speed unless acted upon by a net external force. This is why the friction must be exactly counteracted by the pulling force for constant speed.

Newton's second law (Force = mass * acceleration) establishes the relationship between an object's mass, the net force acting on it, and its acceleration: \[ F = ma \]
To accelerate the wagon, the pulling force must not only overcome friction but also provide additional force to manage the net effect as delineated by the second law.
  • Objects accelerate when a net force is applied.
  • Force and acceleration are directly proportional.
  • Mass and acceleration are inversely proportional in this context.
By understanding these laws, one can predict how the wagon will behave under different force scenarios, as calculated in steps for constant speed and acceleration in the solution.

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Most popular questions from this chapter

Two forces act on a point object as follows: \(100 \mathrm{~N}\) at \(170.0^{\circ}\) and \(100 \mathrm{~N}\) at \(50.0^{\circ}\). Find their resultant.

An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon? (c) What will be its acceleration on the Moon under the action of a \(0.500-\mathrm{N}\) resultant force?

A force of \(100 \mathrm{lb}\) acting on a body weighing \(500 \mathrm{lb}\) causes the body to accelerate uniformly. What would happen to the acceleration if the force is increased to \(200 \mathrm{lb}\) ? [Hint: Units are not important here as long as you are consistent.] Assume no friction.

A car whose weight is \(F_{W}\) is on a ramp, which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. 3-6, the car's weight is a force \(\overrightarrow{\mathrm{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathrm{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

The only force acting on a \(5.0\) -kg object has components \(F_{x}=20 \mathrm{~N}\) and \(F_{y}=30 \mathrm{~N}\). Find the acceleration of the object. Use \(\sum F_{x}=m a_{x}\) and \(\sum F_{y}=m a_{y}\) to obtain $$ \begin{array}{l} a_{x}=\frac{\sum F_{x}}{m}=\frac{20 \mathrm{~N}}{5.0 \mathrm{~kg}}=4.0 \mathrm{~m} / \mathrm{s}^{2} \\ a_{y}=\frac{\sum F_{y}}{m}=\frac{30 \mathrm{~N}}{5.0 \mathrm{~kg}}=6.0 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$ These components of the acceleration are shown in Fig. \(3-10 .\) From the figure, $$ a=\sqrt{(4.0)^{2}+(6.0)^{2}} \mathrm{~m} / \mathrm{s}^{2}=7.2 \mathrm{~m} / \mathrm{s}^{2} $$ and \(\theta=\arctan (6.0 / 4.0)=56^{\circ}\).
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