Compute Torque About Point A - 3 SF

  • Thread starter Thread starter Peter2288
  • Start date Start date
  • Tags Tags
    Torque
Click For Summary

Homework Help Overview

The discussion revolves around calculating torque about a specific point (point A) for three forces, with a focus on the third force (F3). Participants are exploring the relationship between force, distance, and angle in the context of torque calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for torque and question the appropriate distance (r) and angle (θ) to use in calculations. There are inquiries about whether to use the hypotenuse of a triangle for r and how to break down the force into components.

Discussion Status

The discussion is active, with participants providing guidance on the use of angles and components in torque calculations. There is a mix of interpretations regarding the setup and calculations, but no explicit consensus has been reached.

Contextual Notes

Participants are working under the constraint of using three significant figures for their calculations, and there is an emphasis on understanding the geometric relationships involved in the problem.

Peter2288
Messages
5
Reaction score
0
[SOLVED] Torque Question

Homework Statement



See diagram for problem

additional instructions
--------------------
For each of the three forces shown, compute the moment (N m, + CCW) about point A. Assume 3 SF for all numbers.

I am trying to solve for the moment(torque) of F3


Homework Equations



Force * (Distance between force and point A) * sine(theta)??


The Attempt at a Solution



too many to name.
 

Attachments

  • diagram.jpg
    diagram.jpg
    8.6 KB · Views: 441
Physics news on Phys.org
Peter2288 said:

Homework Equations



Force * (Distance between force and point A) * sine(theta)??




That's it! Problem over right there. Tounet = F.r.sinX. Add the torques separately, one by one.
 
For force 3, the r value is supposed to be the direct line between it and the point of pivot right? Do I use the hypotenuse of the triangle for the r value? Also what angle do I use (arctan (3 / 4)??), and am I supposed to somehow break up the force into x and y components? I've tried some of this stuff, but none of it seemed to work. Thanks for your time.
 
Last edited:
Peter2288 said:
For force 3, the r value is supposed to be the direct line between it and the point of pivot right?
Right.
Do I use the hypotenuse of the triangle for the r value?
Sure.
Also what angle do I use (arctan (3 / 4)??),
When using [itex]r F \sin\theta[/itex], the angle is the angle between the position vector and the force vector.
and am I supposed to somehow break up the force into x and y components?
You can always break a force into its components and then find the torque for each component and add them. (That would be much quicker, especially if you realize that [itex]r \sin\theta[/itex] is the perpendicular distance to the line of the force.)
 
Thanks guys.
 

Similar threads

Replies
30
Views
5K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
17
Views
3K
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K