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Chapter 3: Problem 63

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) For constant speed, the force is \(~ 34.64 \, \text{N}\). (b) With acceleration, the force is \(~ 44.06 \, \text{N}\).

Step by step solution

01

Understand the Problem

We have a wagon of mass \( m = 20 \, \text{kg} \) being pulled by a rope inclined at \( 30^\circ \). There is a friction force opposing the motion of \( 30 \, \text{N} \). We need to find the magnitude of the pulling force \( F \) for two cases: (a) constant speed (means acceleration is \( 0 \)) and (b) acceleration \( a = 0.40 \, \text{m/s}^2 \).
02

Analyze Forces when Moving at Constant Speed

For constant speed, acceleration \( a = 0 \). This means the net force acting on the wagon is zero. Vertical forces balance each other, so focus on horizontal forces:\[ F \cos(30^\circ) = F_f \]where \( F_f = 30 \, \text{N} \) is the friction force. Solve for \( F \):\[ F = \frac{F_f}{\cos(30^\circ)} = \frac{30}{\sqrt{3}/2} = 20\sqrt{3} \, \text{N} \]\(~ 34.64 \, \text{N} \).
03

Analyze Forces when Wagon is Accelerating

For acceleration \( a = 0.40 \, \text{m/s}^2 \), use Newton's second law: \( F_{net} = ma \).\[ F \cos(30^\circ) - F_f = ma \]Solve for \( F \):\[ F = \frac{F_f + ma}{\cos(30^\circ)} = \frac{30 + 20 \times 0.40}{\sqrt{3}/2} = \frac{38}{\sqrt{3}/2} = 20 + 8\sqrt{3} \, \text{N} \] \(~ 44.06 \, \text{N} \).

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Key Concepts

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

Friction Force
Friction is a force that resists the motion of objects or surfaces sliding against each other. It opposes the direction of motion and arises due to interactions between the surfaces in contact. In this exercise, the friction force is given as 30 N, which acts opposite to the pulling force of the wagon.

Key aspects of friction to consider:
  • **Magnitude**: The force of friction depends on factors like the nature of the surfaces in contact and the normal force (perpendicular force) pressing them together.
  • **Direction**: Friction acts parallel to the surfaces in contact and opposite the direction of motion.
  • **Types**: The most common types include static friction, which prevents motion, and kinetic friction, which acts when an object is moving.
In our problem, the given friction force is kinetic, since the wagon is in motion. This force must be overcome by the pulling force for any movement to occur, whether it's at constant speed or with acceleration.
Constant Speed
When an object moves at constant speed, its velocity remains unchanged, meaning there's no acceleration. According to Newton's first law of motion, this implies that the net external force acting on the object is zero.
For the wagon moving at constant speed, the forces in the horizontal direction must balance. The friction force of 30 N is exactly balanced by the horizontal component of the pulling force. This is why:
  • **Constant Speed**: Requires that the sum of all forces in the direction of motion adds up to zero.
  • **Calculation**: Using trigonometry, since the rope is inclined, the horizontal component of the pulling force is calculated as \( F \cos(30^\circ) = 30 \, \text{N} \).
  • **Result**: Solving this, we find that the pulling force is approximately 34.64 N.
So, to maintain a constant speed, the applied force must match the friction force in magnitude, overcoming it exactly to result in a net force of zero.
Acceleration
Acceleration occurs when there is a change in velocity, which means a net force acts on the object. According to Newton's second law of motion, the net force on an object is the product of its mass and acceleration.
For the wagon experiencing an acceleration of 0.40 m/s虏, the pulling force must overcome both the friction force and also provide the additional force needed to accelerate the mass of the wagon. Here's how this works:
  • **Newton's Second Law**: States that the net force acting on an object is \( F_{net} = ma \), where \( m \) is the mass and \( a \) is the acceleration.
  • **Calculation**: The horizontal component of the pulling force must be greater than the friction force to account for the net force required for acceleration. We use:
    \[ F \cos(30^\circ) - 30 = 20 imes 0.40 \]
  • **Result**: Solving the above equation, the required pulling force turns out to be approximately 44.06 N.
Thus, to achieve acceleration, the pulling force must not only balance the friction but also provide additional force proportional to the desired acceleration.

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Most popular questions from this chapter

The coefficient of static friction between a box and the flat bed of a truck is \(0.60\). What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{\mathrm{f}}=\mu_{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck: otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{\mathrm{f}}=m a_{x}\) However, if the box is on the verge of slipping, \(F_{\mathrm{f}}=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

A cord passing over a frictionless, massless pulley has a \(4.0\) -kg object tied to one end and a \(12-\mathrm{kg}\) object tied to the other. Compute the acceleration and the tension in the cord.

A \(12.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(700 \mathrm{~m} / \mathrm{s}\) as it travels \(20.0 \mathrm{~cm}\) in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? [Hint: Be careful of units.]

Two cannonballs that each weigh \(4.00 \mathrm{kN}\) on Earth are floating in space far from any other objects. Determine the mutually attractive gravitational force acting on them when they are separated, center-to-center, by \(10.0 \mathrm{~m}\).
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