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Chapter 6: Problem 54

A worker drags a crate across a factory floor by pulling on a rope tied to the crate (Fig. 6-60). The worker exerts a force of \(450 \mathrm{~N}\) on the rope, which is inclined at \(38^{\circ}\) to the horizontal, and the floor exerts a horizontal force of \(125 \mathrm{~N}\) that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is \(310 \mathrm{~kg}\) or \((\mathrm{b})\) its weight is \(310 \mathrm{~N}\).

Short Answer

Expert verified
For (a), acceleration is about 0.739 m/s\textsuperscript{2}. For (b), acceleration is about 7.25 m/s\textsuperscript{2}.

Step by step solution

01

- Understand the problem

The problem involves calculating the acceleration of a crate being dragged with a force applied at an angle, considering opposing frictional force, for two different scenarios: when given mass and when given weight.
02

- Identify given values

Force applied, \( F = 450 \text{ N} \); Angle of force, \( \theta = 38^\text{掳} \); Opposing frictional force, \( F_f = 125 \text{ N} \). For (a): \( \text{mass} = 310 \text{ kg} \) and for (b): \( \text{weight} = 310 \text{ N} \).
03

- Resolve the applied force

Calculate the horizontal component of the applied force: \[ F_x = F \times \text{cos}(\theta) = 450 \text{ N} \times \text{cos}(38^\text{掳}) \approx 354.1 \text{ N} \].
04

- Calculate net force

Determine the net force acting on the crate: \[ F_{\text{net}} = F_x - F_f = 354.1 \text{ N} - 125 \text{ N} = 229.1 \text{ N} \].
05

- Calculate acceleration for mass

Using Newton's Second Law, \( F = ma \), solve for acceleration \( a \): \[ a = \frac{F_{\text{net}}}{m} = \frac{229.1 \text{ N}}{310 \text{ kg}} \approx 0.739 \text{ m/s}^2 \].
06

- Convert weight to mass for (b)

Given the weight of the crate, use \( W = mg \) to find mass: \[ m = \frac{W}{g} = \frac{310 \text{ N}}{9.8 \text{ m/s}^2} \approx 31.6 \text{ kg} \].
07

- Calculate acceleration for weight

Using Newton's Second Law, \( F = ma \), solve for acceleration \( a \) for (b): \[ a = \frac{F_{\text{net}}}{m} = \frac{229.1 \text{ N}}{31.6 \text{ kg}} \approx 7.25 \text{ m/s}^2 \].

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Key Concepts

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

Force resolution
When dealing with forces applied at an angle, it's crucial to break them down into their horizontal and vertical components.
In this exercise, the worker exerts a force of 450 N on a rope inclined at 38 degrees to the horizontal.
The horizontal component of this force is what moves the crate, while the vertical component typically does not contribute to the horizontal motion because the crate is moving on a flat surface.
To resolve the force, we use trigonometric functions:
  • Horizontal component: \( F_x = F \times \cos(\theta) \)
  • Vertical component: \( F_y = F \times \sin(\theta) \)
In our case: \[ F_x = 450 \text{ N} \times \cos(38^\circ) \approx 354.1 \text{ N} \]. This resolved force is essential for subsequent calculations of net force and acceleration.
Frictional force
The frictional force opposes the motion of the crate across the floor.
This is given as 125 N in the exercise.
Frictional force can be calculated using the coefficient of friction between surfaces times the normal force, but in this exercise, we are directly given its value.
The net force acting on the crate is then found by subtracting the frictional force from the horizontal component of the applied force:
  • \( F_{\text{net}} = F_x - F_f \)
  • \( F_{\text{net}} = 354.1 \text{ N} - 125 \text{ N} = 229.1 \text{ N} \)
This net force is used to calculate the acceleration of the crate.
Acceleration calculation
Newton's Second Law is pivotal here, stating that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass:
  • \( F = ma \)
To find the acceleration, we rearrange this formula:\( a = \frac{F_{\text{net}}}{m} \)For part (a) of the exercise, where the mass is given as 310 kg, the acceleration is:
  • \( a = \frac{229.1 \text{ N}}{310 \text{ kg}} \approx 0.739 \text{ m/s}^2 \)
In part (b), the weight of the crate is given as 310 N.
To find the mass from the weight, we use:
  • \( W = mg \)
  • \( m = \frac{W}{g} = \frac{310 \text{ N}}{9.8 \text{ m/s}^2} \approx 31.6 \text{ kg} \)
The acceleration then is:
  • \( a = \frac{229.1 \text{ N}}{31.6 \text{ kg}} \approx 7.25 \text{ m/s}^2 \)
This breakdown ensures you understand how mass and net force influence acceleration.
By mastering these calculations, you can apply Newton's Second Law to various physical scenarios.

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Most popular questions from this chapter

Suppose the coefficient of static friction between the road and the tires on a Formula One car is \(0.6\) during a Grand Prix auto race. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

A student of weight \(667 \mathrm{~N}\) rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force \(\vec{N}\) on the student from the seat is \(556 \mathrm{~N}\). (a) Does the student feel "light" or "heavy" there? (b) What is the magnitude of \(\vec{N}\) at the lowest point? (c) What is the magnitude \(N\) if the wheel's speed is doubled?

A worker wishes to pile a cone of sand onto a circular area in his yard. The radius of the circle is \(R\), and no sand is to spill onto the surrounding area (Fig. 6-46). If \(\mu^{\text {stat }}\) is the static coefficient of friction between each layer of sand along the slope and the sand beneath it (along which is might slip), show that the greatest volume of sand that can be stored in this manner is \(\pi \mu^{\text {stat }} R^{3} / 3\). (The volume of a cone is \(A h / 3\), where \(A\) is the base area and \(h\) is the cone's height.)

In the early afternoon, a car is parked on a street that runs down a steep hill, at an angle of \(35.0^{\circ}\) relative to the horizontal. Just then the coefficient of static friction between the tires and the street surface is \(0.725 .\) Later, after nightfall, a sleet storm hits the area, and the coefficient decreases due to both the ice and a chemical change in the road surface because of the temperature decrease. By what percentage must the coefficient decrease if the car is to be in danger of sliding down the street?

A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction is \(0.35 .\) (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration?
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