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Chapter 4: Problem 37

A \(1000-\mathrm{kg}\) car is pulling a \(300-\mathrm{kg}\) trailer. Together, the car and trailer have an acceleration of \(2.15 \mathrm{~m} / \mathrm{s}^{2}\) in the positive \(x\)-direction. Neglecting frictional forces on the trailer, determine (a) the net force on the car, (b) the net force on the trailer, (c) the magnitude and direction of the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road.

Short Answer

Expert verified
The net force on the car is \(2795 N\), the net force on the trailer is \(645 N\), the magnitude and direction of the force exerted by the trailer on the car is \(645 N\) in the negative x-direction and the resultant force exerted by the car on the road is \(2150 N\) in the positive x-direction.

Step by step solution

01

Determine the net force on the car

Using Newton's second law, find the net force on the car. Since the car and trailer are accelerating together, their combined mass (\(1000 kg + 300 kg = 1300 kg\)) should be used. Calculate the net force using the formula: \(F = ma\), which gives \(F = 1300 kg * 2.15 m/s^2 = 2795 N\)
02

Determine the net force on the trailer

The net force on the trailer is its weight multiplied by the acceleration. Therefore, we can use the formula \(F=ma\) where \(m = 300 kg\) and \(a = 2.15 m/s^2\). Substituting these values, we get \(F = 300 kg * 2.15 m/s^2 = 645 N\)
03

Determine the force exerted by the trailer on the car

The force exerted by the trailer on the car is equal in magnitude and opposite in direction to the force the car exerts on the trailer. Hence, the force exerted by the trailer on the car is also \(645 N\), but in the negative x-direction.
04

Determine the resultant force exerted by the car on the road

The resultant force is the sum of the net force on the car and the force exerted by the trailer on the car. Therefore, the resultant force = \(2795 N - 645 N = 2150 N\) in the positive x-direction.

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Key Concepts

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

Net Force
Net force is the total force acting on an object when all individual forces are combined. It plays a crucial role in determining how an object moves and accelerates.
To calculate the net force, we use Newton's Second Law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This can be represented by the formula:
  • \( F = ma \)
In the exercise, the car and trailer move together with an acceleration, so their masses are combined to determine the net force.
Once we know the acceleration and the total mass, we can compute the net force guiding their motion.
Acceleration
Acceleration is a measure of how quickly an object changes its velocity. It's a vector, which means it has direction as well as magnitude.
In this exercise, the car and trailer have an acceleration of \(2.15 \mathrm{~m} / \mathrm{s}^{2}\) in the positive x-direction.
Acceleration occurs due to a net force acting, as explained by Newton鈥檚 Second Law. Here's how:
  • The greater the net force on an object, the greater its acceleration.
  • Conversely, the larger the mass, the smaller the acceleration for a given force.
Understanding these principles allows us to see why different objects accelerate differently under similar forces.
Force Exertion
Force exertion refers to how much push or pull one object applies to another. It follows Newton's Third Law, which states that for every action, there's an equal and opposite reaction.
In this scenario, the trailer exerts a force on the car. The magnitude of this force is \(645 N\), in the opposite direction to the car's motion. This reaction happens because the car is pulling the trailer, so they act upon each other with equal forces.
The balancing of these exerted forces helps in understanding how connected objects impact each other's motion.
Mass and Motion
The relationship between mass and motion is fundamental in physics. The mass of an object affects how it reacts to forces and moves through space.
For the car and trailer system:
  • The total mass is important for calculating the net force and resultant acceleration.
  • The mass of each individual object (car and trailer) is used to determine specific forces acting on them separately.
This exercise effectively illustrates how both mass and force influence motion, allowing us to predict how an object will behave when subjected to various forces.

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Most popular questions from this chapter

A crate of mass \(m=\) \(32 \mathrm{~kg}\) rides on the bed of a truck attached by a cord to the back of the cab as in Figure \(\mathrm{P} 4.26 .\) The cord can withstand a maximum tension of \(68 \mathrm{~N}\) before breaking. Neglecting friction between the crate and truck bed, find the maximum acceleration the truck can have before the cord breaks.

A fisherman poles a boat as he searches for his next catch. He pushes parallel to the length of the light pole, exerting a force of \(240 \mathrm{~N}\) on the bottom of a shallow lake. The pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of \(35.0^{\circ}\) with the vertical and the water exerts a horizontal drag force of \(47.5 \mathrm{~N}\) on the boat, opposite to its forward velocity of magnitude \(0.857 \mathrm{~m} / \mathrm{s}\). The mass of the boat including its cargo and the worker is \(370 \mathrm{~kg}\). (a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force. (b) Assume the forces are constant over a short interval of time. Find the velocity of the boat \(0.450 \mathrm{~s}\) after the moment described. (c) If the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same, what would happen to buoyant force and the acceleration of the boat?

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

A dockworker loading crates on a ship finds that a \(20-\mathrm{kg}\) crate, initially at rest on a horizontal surface, requires a \(75-\mathrm{N}\) horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of \(60 \mathrm{~N}\) is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

A \(75-\mathrm{kg}\) man standing on a scale in an elevator notes that as the elevator rises, the scale reads \(825 \mathrm{~N}\). What is the acceleration of the elevator?
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