Solving Beam Hinge Torque for Horizontal Force

Click For Summary
SUMMARY

The discussion focuses on calculating the horizontal component of the force exerted by a hinge on a beam inclined at 14.0 degrees, supported by a cable. The beam has a mass of 35.0 kg, and the tension in the cable is analyzed using torque equations. The key equation derived is Tsin(theta) = (1/2)mgcos(theta), where T is the tension, m is the mass, and g is the acceleration due to gravity. Participants emphasize the importance of drawing a complete free body diagram to avoid mistakes in resolving forces and torques.

PREREQUISITES
  • Understanding of torque and force equilibrium principles
  • Familiarity with free body diagrams
  • Knowledge of trigonometric functions in physics
  • Ability to apply Newton's laws of motion
NEXT STEPS
  • Study the concept of torque in static equilibrium
  • Learn how to construct and analyze free body diagrams
  • Explore the relationship between tension and angles in physics problems
  • Review Newton's second law as it applies to rotational motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as educators looking for problem-solving strategies in force and torque analysis.

NAkid
Messages
70
Reaction score
0

Homework Statement


A 35.0kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of theta=14.0o with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)

Homework Equations





The Attempt at a Solution


i know that the horizontal force at pivot must be equal and opposite to horizontal force of tension. so for torques i have
LTsin(theta) = (L/2)mgcos(theta)
Ls cancel
so Tsin(theta) = (1/2)mgcos(theta)

but I don't know what I'm solving for exactly.. when I draw the free body diagram I have that F(horiz)=Fcos(theta), which seems to contradict my equations... help please!
 

Attachments

  • prob19a_beamhinge1.gif
    prob19a_beamhinge1.gif
    1.8 KB · Views: 704
Physics news on Phys.org
NAkid said:
LTsin(theta) = (L/2)mgcos(theta)

About which point are you taking the moments?

but I don't know what I'm solving for exactly.. when I draw the free body diagram I have that F(horiz)=Fcos(theta),...

What is F?

First label the forces properly. Resolve into horizontal and vertical components. Without a neat diagram, you will keep making mistakes. Then take the torques about a suitable point.
 
Draw a complete diagram!

Hi NAkid!

Shooting Star is right.

You must draw in the force F at the hinge, and an angle (phi, perhaps?) for it.

And the tension force and the weight force.

It's no good keeping half the diagram in your head - "Without a neat diagram, you will keep making mistakes", and even miss some things completely.
NAkid said:
i know that the horizontal force at pivot must be equal and opposite to horizontal force of tension.

Yes, and that's a linear force equation, not a torque, isn't it?

And you haven't written it down, anyway!
so for torques i have
LTsin(theta) = (L/2)mgcos(theta)
Ls cancel
so Tsin(theta) = (1/2)mgcos(theta)

I'm not sure what you've done here.

After you've written the linear force equation, do the torque about the hinge, and that should immediatly give you the value of T.

And then … :smile:
 

Similar threads

Replies
6
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
10
Views
3K
Replies
1
Views
6K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
15K
Replies
8
Views
6K