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Chapter 6: Problem 9

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{~N}\). What is the magnitude of the pulling force exerted by his wife?

Short Answer

Expert verified
The wife's pulling force is approximately 45.06 N.

Step by step solution

01

Understand the Work Done by the Husband and Wife

Work done when pulling the wagon depends on the component of the force that acts in the direction of displacement. It can be calculated using the formula: \[ W = F \cdot d \cdot \cos(\theta) \] where \( W \) is the work done, \( F \) is the magnitude of the force, \( d \) is the displacement, and \( \theta \) is the angle of the force with respect to the horizontal.
02

Calculate the Work Done by the Husband

Since both husband and wife do the same work, express the husband's work as: \[ W_h = F_h \cdot d \cdot \cos(58^{\circ}) \] With \( F_h = 67 \mathrm{~N} \), this becomes: \[ W_h = 67 \cdot d \cdot \cos(58^{\circ}) \]
03

Express the Work Done by the Wife

Express the wife's work in terms of her pulling force \( F_w \): \[ W_w = F_w \cdot d \cdot \cos(38^{\circ}) \] Because they do the same amount of work: \[ W_h = W_w \]
04

Set the Work Equations Equal and Solve for the Wife鈥檚 Force

Since \( W_h = W_w \), equate the expressions: \[ 67 \cdot d \cdot \cos(58^{\circ}) = F_w \cdot d \cdot \cos(38^{\circ}) \] Cancel \( d \) from both sides of the equation: \[ 67 \cdot \cos(58^{\circ}) = F_w \cdot \cos(38^{\circ}) \] Solve for \( F_w \): \[ F_w = \frac{67 \cdot \cos(58^{\circ})}{\cos(38^{\circ})} \]
05

Calculate the Magnitude of the Pulling Force Exerted by the Wife

Substitute the values and compute: - \( \cos(58^{\circ}) \approx 0.5299 \) - \( \cos(38^{\circ}) \approx 0.7880 \) \[ F_w = \frac{67 \cdot 0.5299}{0.7880} \approx 45.06 \mathrm{~N} \] Thus, the wife's force is approximately 45.06 N.

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Key Concepts

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

Force and Motion
In physics, understanding force and motion is crucial. **Force** refers to any interaction that, when unopposed, will change the motion of an object. It can be a push or a pull and is measured in newtons (N). **Motion** is the change in position of an object with respect to time. These concepts are related through Newton's laws of motion.
The exercise considers force's effect on motion 鈥� specifically, how force directed at an angle impacts the work done. Work is done when a force causes an object to displace. The horizontal sidewalk scenario in the exercise is an intuitive way to visualize how force influences motion, considering variables like angle and magnitude.
Trigonometry in Physics
Trigonometry plays a pivotal role in physics by allowing us to resolve forces into components. When a force is applied at an angle, it can be broken down into horizontal and vertical components using trigonometric functions. The horizontal component influences the displacement in force and motion problems.
In the given problem, the husband and wife's forces are at different angles - 58掳 and 38掳, respectively. The cosine function is used here because it helps to find the component of the force that acts in the direction of displacement. Using \[ \text{cos}(\theta) \] gives the ratio of the adjacent side (horizontal force) over the hypotenuse (the force itself) in a right-angled triangle formed by these components. This is why **cosine** is crucial in calculating the work done by each person pulling the wagon.
Physics Problem Solving
Physics problem solving often requires methodical steps to reach a solution. It's crucial to understand the problem's given data and what needs to be determined. This involves identifying relevant formulas and concepts such as work, force, and angles.
In this exercise, the problem is solved by equating the work done by both individuals. Steps involve:
  • Understanding that both do equal amounts of work, given by the formula \[ W = F \cdot d \cdot \cos(\theta) \]
  • Substituting known values to find unknown forces.
  • Using trigonometric calculations to resolve forces.

This methodical approach ensures that all aspects of the physical situation are considered, leading to an insightful and accurate solution. Mastery in problem-solving comes from practice and applying these structured methods consistently.

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Most popular questions from this chapter

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of \(r=36 \mathrm{~m} .\) Neglect friction and air resistance. What must be the height \(h\) of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

You are moving into an apartment and take the elevator to the 6th floor. Does the force exerted on you by the elevator do positive or negative work when the elevator (a) goes up and (b) goes down? Explain your answers. Suppose your weight is \(685 \mathrm{~N}\) and that of your belongings is \(915 \mathrm{~N}\). (a) Determine the work done by the elevator in lifting you and your belongings up to the 6 th floor \((15.2 \mathrm{~m})\) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity? Check to see that your answers are consistent with your answers to the Concept Questions.

One kilowatt hour (kWh) is the amount of work or energy generated when one kilowatt of power is supplied for a time of one hour. A kilowatt hour is the unit of energy used by power companies when figuring your electric bill. Determine the number of joules of energy in one kilowatt hour.

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by Lance Armstrong \((m=75.0)\) is \(6.50 \mathrm{~W}\) per kilogram of his body mass. (a) How much work does he do during a \(135-\mathrm{km}\) race in which his average speed is \(12.0 \mathrm{~m} / \mathrm{s} ?\) (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.

A \(5.00 \times 10^{2}-\) kg hot-air balloon takes off from rest at the surface of the earth. The nonconservative wind and lift forces take the balloon up, doing \(+9.70 \times 10^{4} \mathrm{~J}\) of work on the balloon in the process. At what height above the surface of the earth does the balloon have a speed of \(8.00 \mathrm{~m} / \mathrm{s} ?\)
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