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Chapter 9: Problem 28

A man drags a 72 -kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted \(25^{\circ}\) above the horizontal, and the strap is inclined \(61^{\circ}\) above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap.

Short Answer

Expert verified
The tension in the strap is approximately 548 N.

Step by step solution

01

Identify Given Information

We have a crate with a mass of 72 kg. It is tilted at an angle of \(25^{\circ}\), and the strap is inclined \(61^{\circ}\) above the horizontal. We want to find the tension in the strap that keeps the crate moving at a constant velocity.
02

Draw the Free Body Diagram

In the diagram, illustrate all the forces acting on the crate: the gravitational force \(mg\), the normal force \(N\), the frictional force \(f\), and the tension \(T\) in the strap. The gravitational force acts downward, the normal force acts perpendicular to the surface of the crate, the frictional force opposes motion, and the tension is directed along the strap at \(61^{\circ}\).
03

Apply Newton's First Law

Since the crate moves at a constant velocity, all forces must be balanced. This means the net force in both the horizontal and vertical directions is zero. We'll apply Newton's First Law: \(\Sigma F = 0\).
04

Resolve Forces into Components

The tension \(T\) can be resolved into horizontal and vertical components. The horizontal component \(T_x = T \cos(61^{\circ})\) and the vertical component \(T_y = T \sin(61^{\circ})\). Similarly, the gravitational force \(mg\) can be resolved where \(mg_x = mg \sin(25^{\circ})\) and \(mg_y = mg \cos(25^{\circ})\).
05

Set Up Equations for Balanced Forces

In the horizontal direction: \(T \cos(61^{\circ}) = f\).In the vertical direction: \(T \sin(61^{\circ}) + N = mg \cos(25^{\circ})\). The frictional force \(f = \mu N\) where \(\mu\) is the coefficient of friction. Substituting for \(f\) in terms of \(T\) and solving for \(T\).
06

Solve for Tension T

With all previous equations set up, solve the system: 1. Find \(N\) from the vertical balance: \(T \sin(61^{\circ}) + N = mg \cos(25^{\circ})\).2. Use \(f = T \cos(61^{\circ})\), insert into \(f = \mu N\) to find \(\mu\).3. Solve the equations to find the value of tension \(T\) by equating \(T \cos(61^{\circ})\) with the expression for friction \(\mu N\).
07

Substitute Values and Calculate

Substitute known values: \(m = 72\, \mathrm{kg}\), \(g = 9.8\, \mathrm{m/s^2}\), and solve for \(T\) using the trigonometric identities. Perform calculations ensuring consistent units. Confirm that tension balances the forces for no acceleration.

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Key Concepts

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

Free Body Diagram
A free body diagram is a simple drawing that displays all external forces acting on an object. In the context of our crate problem, this diagram is crucial for visualizing how forces interact when the crate is dragged across the floor.

In our situation, the key forces include:
  • **Gravitational Force (\(mg\))**: This force pulls the crate downward due to gravity.
  • **Normal Force (\(N\))**: This is a perpendicular force exerted by the floor, supporting the crate.
  • **Frictional Force (\(f\))**: Opposes the crate's motion, acting along the floor.
  • **Tension (\(T\)):** Directed along the strap, inclined at \(61^{\circ}\) relative to the horizontal direction.
When constructing a free body diagram, accurately marking the direction and relative size of each force helps in applying Newton's laws effectively. This visual aid helps in understanding how the forces are balanced when the crate moves with constant velocity.
Tension Force
The tension force in the strap is the pulling force transmitted through the strap, which helps in dragging the crate. Understanding the tension force is key to solving for its magnitude, as the strap determines how the pull impacts other forces.

Since the strap is inclined at \(61^{\circ}\) to the horizontal, tension isn't just a single push or pull. It's resolved into two components:
  • **Horizontal Component (\(T_x = T \cos(61^{\circ})\)):** This helps counteract the frictional force.
  • **Vertical Component (\(T_y = T \sin(61^{\circ})\)):** Works against part of the gravitational pull together with the normal force.
This resolution into components is essential for applying Newton's First Law effectively, ensuring that all forces are balanced and the crate remains in motion without acceleration.
Frictional Force
The frictional force is a resistive force that opposes the motion of the crate across the floor. It's an essential aspect of the problem as it should be balanced by the horizontal component of tension when the crate moves at a constant speed.

Frictional force is calculated using:
  • **Friction Equation:** \(f = \mu N\), where \(\mu\) is the coefficient of friction, which depends on the surfaces in contact.
  • **Horizontal Balance:** For constant velocity, \(T \cos(61^{\circ}) = f\)
Determining the values for normal force and \(\mu\) is important to quantify the frictional force. Once you've computed these, it ensures an understanding of how tension in the strap counteracts this resistive force efficiently.
Force Components
Force components are simply parts of a force acting in specific directions. Breaking down forces into components along the horizontal and vertical axes makes analyzing the system easier, especially when forces act at angles.

For the crate:
  • **Tension Components:** * \(T_x = T \cos(61^{\circ})\) for horizontal * \(T_y = T \sin(61^{\circ})\) for vertical.
  • **Gravity Components:** * \(mg_x = mg \sin(25^{\circ})\) not primarily needed here but illustrates tilt effects * \(mg_y = mg \cos(25^{\circ})\) for balance against vertical supports.
    • By resolving forces into components, identifying how they interact with each other becomes straightforward. This method aids in setting up equations that reflect Newton's First Law, ensuring equilibrium in all directions during constant velocity motion.

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