Static Equilibrium Cable Problem

Click For Summary
SUMMARY

The discussion focuses on solving a static equilibrium problem involving a 50.0 kg uniform square sign suspended from a 3.00 m horizontal rod. The cable attached to the rod extends to a wall 4.00 m above the hinge point. The key equations applied are net force and net torque equating to zero. The tension in the cable and the horizontal and vertical components of the force exerted by the wall on the rod are the primary calculations required to solve the problem.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of torque calculations
  • Familiarity with free body diagrams
  • Basic physics of forces and tension
NEXT STEPS
  • Study the concept of torque in depth, focusing on its application in static systems
  • Learn how to construct and analyze free body diagrams for complex structures
  • Investigate the principles of tension in cables and their role in static equilibrium
  • Explore real-world applications of static equilibrium in engineering and architecture
USEFUL FOR

This discussion is beneficial for physics students, engineering students, and professionals involved in structural analysis or mechanics, particularly those dealing with static systems and equilibrium problems.

Seraph404
Messages
63
Reaction score
0

Homework Statement



PhysProb20Ch13.jpg


In fig. 13-32, a 50.0 kg uniform square sign, 2.00 m on a side, is hung from a 3.00 m horizantal rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.00 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? What are the magnitudes and directions of the (b) horizantal and (c) vertical components of the force on the rod from the wall?


Homework Equations



Net force = zero
Net torque = zero


The Attempt at a Solution



I don't understand how there is a vertical component exerted by the wall onto the rod.
 
Last edited:
Physics news on Phys.org
If the weight of the sign was all concentrated at the right corner, there would be no vert comp of the wall on the rod; however, the sign's resultant weight acts thru its c.g., not at the corner. Try summing torques about the top left corner.
 
All right. Thanks so much!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 57 ·
2
Replies
57
Views
7K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
8
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K