# PHYS 2211 Module 11.5

## Examples of Static Equilibrium

11.5 Examples of Static Equilibrium

### Learning Objectives

By the end of this section, you will be able to:

• Identify and analyze static equilibrium situations
• Set up a free-body diagram for an extended object in static equilibrium
• Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations

Practice!

Torque is calculated by multiplying the perpendicular components of the force and the distance from the origin to where the force is applied. The force of gravity on the beam, the beam’s weight, is its mass times the acceleration due to gravity: w = mg = (150 kg)(9.8 m/s2) = 1470 N. The torque due to the weight of the beam, measured with respect to the hinge, is then = (1470)(2 m) = 2940 N•m.

For a beam in static equilibrium, the net torque about the hinge is zero: , where torque is calculated with or . Measured with respect to the hinge, the torque from the cable is and the torque from the weight of the beam is , where the lever arm is 2 m because the force of gravity acts on the beam at its center of mass. When in static equilibrium, these two torques must be equal. So equate them and solve for the force of tension: = 1620 N.

For a beam in static equilibrium, the net torque about the hinge is zero: , where torque is calculated with or . Measured with respect to the hinge, the torque from the cable is and the torque from the weight of the beam is , where the lever arm is 2 m because the force of gravity acts on the beam at its center of mass, and the torque from the hanging mass is . When in static equilibrium, the two clockwise torques must be equal to the counterclockwise torque: . With this equation you can solve for the tension force: = 2830 N.

Since the beam is in static equilibrium, the net force on the beam is also zero. In the vertical (y) direction: = 0. Solving for the vertical component of the force exerted on the beam by the hinge: = (150 kg)(9.8 m/s2) + (75 kg)(9.8 m/s2) – (2833 N)(sin 27°) = 920 N.

Discuss!

The automatic flag raising system on a horizontal flagpole attached to the vertical outside wall of a tall building has become stuck. The management of the building wants to send a person crawling out along the flagpole to fix the problem. Because of your physics knowledge, you have been asked to consult with a group to decide whether or not this is possible.

You are all too aware that no one could survive the 250 foot fall from the flagpole to the ground. The flagpole is a 120-lb steel I-beam which is very strong and rigid. One side of the flagpole is attached to the wall of the building by a hinge so that it can rotate vertically. Nine feet away, the other end of the flagpole is attached to a strong, lightweight steel cable. The cable goes up from the flagpole at an angle of 30° until it reaches the building where it is bolted to the wall. The mechanic who will climb out on the flagpole weighs 150 lbs including equipment.

From the specifications of the building construction, both the bolt attaching the cable to the building and the hinge have been tested to hold a maximum force of 500 lbs. Your boss has decided that the worst case scenario is when the mechanic is at the far end of the flagpole, nine feet from the building.

Should you allow the mechanic to climb out to the end of the pole to make the repairs?