Why is my torque calculation incorrect for this simple force problem?

  • Thread starter Thread starter VinnyCee
  • Start date Start date
  • Tags Tags
    Force
Click For Summary
SUMMARY

The discussion centers on the incorrect torque calculation in a force problem involving a force vector and a position vector. The user calculated the force components as \(\overrightarrow{F_{A_x}} = 375\sin(30^\circ)\) N and \(\overrightarrow{F_{A_y}} = 375\cos(30^\circ)\) N, but encountered issues when computing the torque using the cross product of the position vector \(\overrightarrow{r_{AP}} = (4\hat{i} - 3\hat{j})\) m and the force vector. A key correction noted is that the y-component of the force should be negative, and the position vector must extend from the axis of rotation to the point of force application.

PREREQUISITES
  • Understanding of vector components in physics
  • Knowledge of torque calculation using cross products
  • Familiarity with trigonometric functions and their application in force resolution
  • Basic understanding of coordinate systems in mechanics
NEXT STEPS
  • Study the principles of torque and rotational dynamics
  • Learn about vector cross products in physics
  • Review trigonometric identities and their applications in force analysis
  • Explore examples of force and torque problems in mechanics
USEFUL FOR

Students and professionals in physics, mechanical engineering, and anyone involved in solving problems related to forces and torque calculations.

VinnyCee
Messages
486
Reaction score
0
http://img403.imageshack.us/img403/879/problem4998tl.jpg

Here I resolved the forces:

\overrightarrow{F_{A_x}} = \left(375sin30\right)N

\overrightarrow{F_{A_y}} = \left(375cos30\right)N

Here is the r vector (I think):

\overrightarrow{r_{AP}} = \left(4\hat{i} - 3\hat{j}\right)m

But when I do r X F, I get something wrong, what do I have to do to solve this simple problem?
 
Last edited by a moderator:
Physics news on Phys.org
Let the Centre of AP be M, then MP is the radius Vector I guess from what I understood after seeing the picture. You know the co-ordinate of A and P, so you can find the co-ordinates of M.
 
VinnyCee said:
http://img403.imageshack.us/img403/879/problem4998tl.jpg

Here I resolved the forces:

\overrightarrow{F_{A_x}} = \left(375sin30\right)N

\overrightarrow{F_{A_y}} = \left(375cos30\right)N

Here is the r vector (I think):

\overrightarrow{r_{AP}} = \left(4\hat{i} - 3\hat{j}\right)m

But when I do r X F, I get something wrong, what do I have to do to solve this simple problem?

The y component fo your force should be negative.
Also, in calculating the torque, the vector r should go from the axis of rotation toward the point where the force is applied.

Patrick
 
Last edited by a moderator:

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
3K
  • · Replies 24 ·
Replies
24
Views
2K
Replies
12
Views
2K
  • · Replies 116 ·
4
Replies
116
Views
7K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
1K