Chapter 8, Problem 29P

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# Figure P8.29 shows a uniform beam of mass m pivoted at its lower end, with a horizontal spring attached between its top end and a vertical wall. The beam makes an angle θ with the horizontal. Find expressions for (a) the distance d the spring is stretched from equilibrium and (b) the components of the force exerted by the pivot on the beam.Figure P8.29

(a)

To determine
The distance d of the spring stretched from equilibrium.

Explanation

Given Info: The mass of the beam is m, horizontal angle of the beam is Î¸ , and distance of the spring is d.

The following figure shows the free body diagram of the uniform beam,

Formula to calculate the spring force by using Hookeâ€™s law is,

Fs=kd (1)

• k is the spring constant.
• d is the distance of the spring.

The net torque is zero at the lower end of the beam. Therefore,

âˆ‘Ï„=0FslsinÎ¸âˆ’mg(l2cosÎ¸)=0

• g is the acceleration due to gravity.
• Fs is the spring force.
• l is the length of the beam.
• Î¸ is the horizontal angle.
• m is the mass of the beam

(b)

To determine
The components of force exerted by the pivot on the beam.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started