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Chapter 4: Problem 70

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of \(+11 \mathrm{~m} / \mathrm{s}\) in a time of \(28 \mathrm{~s}\). The combined mass of the boat and trailer is 410 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car?

Short Answer

Expert verified
The tension in the hitch is 161.13 N.

Step by step solution

01

Identify Known Values

First, identify the known values from the problem statement. The initial velocity (\(u\)) is \(0 \, \text{m/s}\), the final velocity (\(v\)) is \(+11 \, \text{m/s}\), the time (\(t\)) is \(28 \, \text{s}\), and the mass (\(m\)) of the boat and trailer is \(410 \, \text{kg}\).
02

Calculate Acceleration

Use the formula for acceleration: \[ a = \frac{v - u}{t} \]Substitute the known values: \[ a = \frac{11 \, \text{m/s} - 0 \, \text{m/s}}{28 \, \text{s}} = \frac{11}{28} \approx 0.393 \, \text{m/s}^2 \]
03

Calculate the Tension in the Hitch

Apply Newton鈥檚 second law of motion, \( F = ma \), to find the tension in the hitch. Since the frictional force is ignored, the tension is the only horizontal force acting on the trailer, which provides the acceleration calculated in Step 2. \[ F = 410 \, \text{kg} \times 0.393 \, \text{m/s}^2 = 161.13 \, \text{N} \]
04

Conclusion

The tension in the hitch that connects the trailer to the car is \(161.13 \, \text{N}\). This force is responsible for accelerating the boat and trailer to the needed velocity.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that explains how the motion of an object changes when it's acted upon by a force. The law is expressed in the equation \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced. This relationship implies that the acceleration of an object is directly proportional to the net force acting upon it, and inversely proportional to its mass.

In the context of the exercise problem, the tension in the hitch represents the force \( F \) that needs to be calculated. By applying this law, the tension is determined as the product of the total mass of the boat and trailer and their acceleration. This calculation is possible because all forces, except the tension, have been either negated or ignored (e.g., friction), simplifying the equation to where the tension alone is responsible for the acceleration.
Kinematics
Kinematics refers to the study of motion without considering the forces that cause it. It involves understanding various parameters like velocity, acceleration, displacement, and time.

For the given problem, kinematics helps in determining the acceleration of the boat and trailer. Given the initial velocity (\( u = 0 \)), final velocity (\( v = 11 \) m/s), and time (\( t = 28 \) s), we use the kinematic formula for acceleration: \[ a = \frac{v - u}{t} \] Substituting the known values gives us the acceleration as \( 0.393 \) m/s虏. This step is crucial as it allows us to link the concepts of motion and forces in further calculations involving Newton鈥檚 Second Law.
Force Calculation
Force calculation is about figuring out the size of the force needed to produce a specific motion. With the use of Newton's Second Law, we can calculate the force required to achieve an acceleration for a given mass.

In solving this problem, once acceleration is determined from kinematic equations, we move on to calculate the required force, or in this case, the tension in the hitch. Using the formula \( F = ma \), we substitute the known values: the mass \( m = 410 \) kg and the previously calculated acceleration \( a = 0.393 \) m/s虏.

The resulting force \( F = 161.13 \) N, represents the tension in the hitch, meaning it's the force exerted by the hitch to accelerate the boat and trailer to the desired velocity. This process illustrates how forces needed in practical situations can be calculated accurately using these fundamental physics principles.

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Most popular questions from this chapter

A \(1.40-\mathrm{kg}\) bottle of vintage wine is lying horizontally in the rack shown in the drawing. The two surfaces on which the bottle rests are \(90.0^{\circ}\) apart, and the right surface makes an angle of \(45.0^{\circ}\) with respect to the ground. Each surface exerts a force on the bottle that is perpendicular to the surface. What is the magnitude of each of these forces?

The central ideas in this problem are reviewed in Multiple-Concept Example \(9 .\) One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches \(47.0 \mathrm{~N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

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?

A person is attempting to push a refrigerator across a room. He exerts a horizontal force on the refrigerator, but it does not move. (a) What other horizontal force must be acting on the refrigerator? (b) How are the magnitude and direction of this force related to the force that the person exerts? (c) Suppose that the person applies the force such that it is the largest possible force before the refrigerator begins to move. What factors determine the magnitude of this force? Problem Consult Multiple-Concept Example 9 to explore a model for solving this problem. A person pushes on a 57 -kg refrigerator with a horizontal force of \(-267 \mathrm{~N}\); the minus sign indicates that the force is directed along the \(-x\) direction. The coefficient of static friction is \(0.65 .\) (a) If the refrigerator does not move, what is the magnitude and direction of the static frictional force that the floor exerts on the refrigerator? (b) What is the magnitude of the largest pushing force that can be applied to the refrigerator before it just begins to move?

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?
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