Finding Torque in a Square: How to Determine the Distance for Calculating Torque

  • Thread starter Thread starter Redfire66
  • Start date Start date
  • Tags Tags
    Square Torque
Click For Summary
To calculate torque in a square with the pivot at the bottom left corner, the distance used should be determined based on the direction of the applied force. The relevant distance can be either the distance from the pivot to the center of mass or the projection of that distance along the axis of the force. It's essential to visualize the force direction on a diagram to clarify which distance value (R or x) is appropriate for torque calculation. The torque formula, T = Fd, emphasizes that the effective distance depends on the angle of the force applied. Understanding the relationship between force direction and distance is crucial for accurate torque calculations.
Redfire66
Messages
36
Reaction score
0

Homework Statement


So image we have a square. And the center of mass is, well in the center.
If I choose the pivot to be the bottom left corner, would the distance used to calculate torque be from the bottom left to the center or to the middle on the same axis? I have a picture since it's kind of hard to explain
Technically I'm just asking, according to the picture, would I use R or x as my distance value for torque

Homework Equations


T = Fd
 

Attachments

  • XR.png
    XR.png
    1.8 KB · Views: 531
Physics news on Phys.org
Torque is the cross product of the displacement vector and the force vector.

So the distance value you want will be some where between zero and 'r'. It depends on the direction of the force.

Draw the direction of the force on your diagram, hopefully which distance value is relevant should become clear..
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 20 ·
Replies
20
Views
3K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
13
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
10K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K