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Chapter 6: Problem 40

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed \(1100 \mathrm{~N}\). The coefficient of static friction between the box and the floor is \(0.35 .\) (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

Short Answer

Expert verified
The angle is found by maximizing the equation. The weight is then calculated using the value of the optimal angle in the given equations.

Step by step solution

01

- Identify Forces

Consider the forces acting on the box: the tension force (T), the friction force (F), the normal force (N), and the gravitational force (mg). The given maximum tension force is 1100 N, and the coefficient of static friction is 0.35.
02

- Resolve Tension Force

Resolve the tension force (T) into its horizontal and vertical components:Horizontal component: \[ T_x = T \, \cos \theta \] Vertical component: \[ T_y = T \, \sin \theta \]
03

- Write the Equations for Equilibrium

For the box to be in equilibrium, the following conditions must be satisfied:Horizontal equilibrium: \[ T_x = F \] Vertical equilibrium: \[ N + T_y = mg \]
04

- Friction Force Equation

The static friction force is given by:\[ F = \mu_s \, N \] Substitute the expression for N from the vertical equilibrium equation:\[ F = \mu_s \, (mg - T \, \sin \theta) \]
05

- Combine Equations

Combine the friction force and horizontal equilibrium equations:\[ T \, \cos \theta = \mu_s \, (mg - T \, \sin \theta) \]Rearrange to solve for \( \theta \): \[ \tan \theta = \frac {T \, \sin \theta}{T \, \cos \theta} = \frac {T}{\mu_s \, mg - T} \]
06

- Solve for Optimal Angle

The angle \( \theta \) that maximizes the amount of sand can be found by setting the derivative of mg with respect to \( \theta \) to zero. By solving this, you can find the optimal angle.
07

- Calculate Weight

Use the value of \( \theta \) found in the previous step to calculate the weight (mg) using the equation obtained earlier:\[ mg = \frac{T \cos \theta}{\mu_s} + T \, \sin \theta \]

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Key Concepts

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

Tension Force
Tension force is a pulling force transmitted along a rope, string, cable, or similar object. In this exercise, we are given a tension force of 1100 N. This tension force must be resolved into its horizontal and vertical components:
Horizontal component: \( T_x = T \cos \theta \)
Vertical component: \( T_y = T \sin \theta \)
We need to ensure these components contribute effectively to overcome the static friction and lift the box.
Equilibrium Conditions
For any object to stay at rest or move with a constant velocity, the equilibrium conditions must be satisfied鈥攎eaning the sum of forces in both the horizontal and vertical directions must be zero.
  • Horizontal equilibrium: The horizontal component of the tension force must be equal to the friction force opposing motion: \( T_x = F \)
  • Vertical equilibrium: The normal force plus the vertical component of the tension force must balance the weight of the box: \( N + T_y = mg \)

Friction Force Calculation
The friction force is the opposing force that resists the relative motion of two surfaces in contact. It is given by:
\[ F = \mu_s \cdot N \]
where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. In this problem, we substitute the normal force expression from the vertical equilibrium equation:
\[ F = \mu_s (mg - T \sin \theta) \]
  • This helps us connect friction to other forces acting on the box, making further calculations for equilibrium possible.

Angle Optimization
Determining the optimal angle of the cable is crucial for maximizing the weight of the sand and box that can be moved. Using the combined expressions for friction and horizontal equilibrium:
\[ T \cos \theta = \mu_s (mg - T \sin \theta) \]
we solve for \( \theta \). This often involves setting the derivative of the weight of the sand to zero to find its maximum value.
This mathematical step identifies the specific angle \( \theta \) that maximizes the amount of sand that can be pulled with the given tension force constraint.
Free Body Diagrams
A Free Body Diagram (FBD) is an essential tool in solving static friction problems. It helps visualize and break down all forces acting on the object. In our scenario, the FBD should include:
  • The gravitational force \( mg \) acting downward
  • The normal force \( N \) acting upward
  • The tension force \( T \) which has both horizontal \( T_x \) and vertical \( T_y \) components
  • The friction force \( F \) opposing the direction of intended motion

Using FBDs to clearly represent these forces helps in setting up the equilibrium equations accurately and understanding how the forces interact.

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Most popular questions from this chapter

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the radius of the circle is \(10 \mathrm{~m}\). What are the magnitude and direction of the force of the boom on the car at the top of the circle if the car's speed there is (a) \(5.0 \mathrm{~m} / \mathrm{s}\) and (b) \(12 \mathrm{~m} / \mathrm{s}\) ?

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?

Suppose you are sitting in a car that is speeding up. Assume the car has rear-wheel drive. (a) Draw free-body diagrams for your own body, the seat in which you are sitting (apart from the car), the car (apart from the seat), and the road surface where the tires and the road interact. (b) Describe each force in words; show larger forces with longer arrows. (c) Identify the third-law pairs of forces. (d) Explain carefully in your own words the origin of the force imparting acceleration to the car.

A spaceship lifts off vertically from the Moon, where the freefall acceleration is \(1.6 \mathrm{~m} / \mathrm{s}^{2} .\) If the spaceship has an upward acceleration of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) as it lifts off, what is the magnitude of the force of the spaceship on its pilot, who weighs \(735 \mathrm{~N}\) on Earth?

We know that as an object passes through the air, the air exerts a resistive force on it. Suppose we have a spherical object of radius \(R\) and mass \(m\). What might the force plausibly depend on? \- It might depend on the properties of the object. The only ones that seem relevant are \(m\) and \(R\). \- It might depend on the object's coordinate and its derivatives: \(\vec{r}, \vec{v}, \vec{a}, \ldots\) \- It might depend on the properties of the air, such as the density, \(\rho\). (a) Explain why it is plausible that the force the air exerts on a sphere depends on \(R\) but implausible that it depends on \(m\). (b) Explain why it is plausible that the force the air exerts depends on the object's speed through it, \(|\vec{v}|\), but not on its position, \(\vec{r}\), or acceleration, \(\vec{a}\). (c) Dimensional analysis is the use of units (e.g., meters, seconds, or newtons) associated with quantities to reason about the relationship between the quantities. Using dimensional analysis, construct a plausible form for the force that air exerts on a spherical body moving through it.
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