Tricky Torque Problem: Solving for Tension and Understanding Beam Forces

  • Thread starter Arooj
  • Start date
  • Tags
    Torque
In summary, the problem involves a massless beam and the conditions for equilibrium, including net torque and net force being zero. The tension and compression forces can be determined by applying these conditions and using trigonometric functions.
  • #1
Arooj
40
0

Homework Statement


65zm9i.jpg


The writing was just my reasoning, which I don't think is right.
Assume the beam is massless.

Homework Equations


T = Fd sin θ
F = mg
net force must be zero.
net torque must be zero.

The Attempt at a Solution


d * 1/2F + 980 = d * (sin 37) T
1/2 F + 980 = (sin 37) T
I don't know what to do from here.
I can mathematically solve the problem to an extent, but conceptually I don't understand how to do the problem. Can someone explain to me how to do the problem from scratch and the reasoning for steps (like what forces are acting on the beam)?
 
Physics news on Phys.org
  • #2
Arooj said:
d * 1/2F + 980 = d * (sin 37) T
1/2 F + 980 = (sin 37) T
What's F? I thought the beam was massless.

To solve this sort of problem, apply the conditions for equilibrium. The net torque = 0 is one of them. What are the others?
 
  • #3
ah, so there is no F, meaning it would just be 980 = sin 37 T, where T is the tension.
net force = zero, forces left = forces right, forces up = forces down, clockwise torques = counterclockwise torques, x and y components of force may separately be set to 0. And T cos 37 would give the force exerted by the wall, I assume?
 
  • #4
Arooj said:
ah, so there is no F, meaning it would just be 980 = sin 37 T, where T is the tension.
Right.
net force = zero, forces left = forces right, forces up = forces down, clockwise torques = counterclockwise torques, x and y components of force may separately be set to 0. And T cos 37 would give the force exerted by the wall, I assume?
Exactly.
 
  • #5
And compression would thus be opposite to the force exerted by the wall, for a net force of zero?
 
  • #6
Arooj said:
And compression would thus be opposite to the force exerted by the wall, for a net force of zero?
Yes, the net force is zero. The force exerted by the wall is the compression.
 

1. What is a "Tricky Torque Problem"?

A "Tricky Torque Problem" refers to a physics problem that involves calculating the torque, or rotational force, acting on an object. These types of problems often require a combination of knowledge in mechanics, mathematics, and problem-solving skills.

2. How do I approach solving a "Tricky Torque Problem"?

First, identify all the known quantities, such as the distance from the pivot point, the force applied, and the angle of rotation. Then, use the formula for torque, T = r x F x sin(theta), to calculate the torque. Finally, check your answer for units and make sure it makes sense in the context of the problem.

3. What are some common mistakes to avoid when solving a "Tricky Torque Problem"?

One common mistake is forgetting to convert units to the correct form, such as converting distance from meters to centimeters. Another mistake is forgetting to use the correct angle of rotation, which should be measured from the direction of the force to the direction of the lever arm.

4. Can you provide an example of a "Tricky Torque Problem" and its solution?

Sure, for example, a wrench is used to loosen a bolt by applying a force of 50 N at a distance of 0.2 m from the pivot point. What is the torque on the bolt?

Using the formula T = r x F x sin(theta), we have T = (0.2 m) x (50 N) x sin(90 degrees) = 10 Nm. Therefore, the torque on the bolt is 10 Nm.

5. How can I improve my skills in solving "Tricky Torque Problems"?

Practice, practice, practice! The more you encounter and solve these types of problems, the more familiar you will become with the concepts and techniques involved. You can also seek help from a teacher or tutor, or use online resources and practice problems to improve your skills.

Similar threads

Replies
6
Views
634
  • Introductory Physics Homework Help
Replies
5
Views
954
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
976
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
596
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top