# Homework Help: A force F is exerted on the top right corner of a rectangular plate at

1. Apr 16, 2013

### cmkc109

1. The problem statement, all variables and given/known data

A force F is exerted on the top right corner of a rectangular plate at an angle 60o
above the horizontal, as shown below. The magnitude of the torque of the force F
about point A, the lower left corner is given by

2. Relevant equations

Both y and x are being pulled upwards, so that is counterclockwise..
but answer is Fx sin 60 + Fy cos 60

3. The attempt at a solution

No idea..

2. Apr 16, 2013

### CWatters

Hint:

#### Attached Files:

• ###### Torque.png
File size:
4 KB
Views:
281
3. Apr 16, 2013

### cmkc109

Thank you!
but I'm still kinda lost...

Last edited: Apr 16, 2013
4. Apr 16, 2013

### cmkc109

one question, where do i put the angle in?

5. Apr 16, 2013

### CWatters

I'll do a new drawing as I don't think my first was very helpful. Back soon.

6. Apr 16, 2013

### cmkc109

ok thanks, check ur msg too please , thanks

7. Apr 16, 2013

### CWatters

Try this drawing. The idea is to split the force F into two component forces. Then multiply each component force by it's distance from the pivot to give two torques (Torque = force * distance). Finally you simply add the two torques.

#### Attached Files:

• ###### Torque.png
File size:
39.2 KB
Views:
319
8. Apr 16, 2013

### cmkc109

this is the drawing i had, but from this picture, it shows that it is Fy sin 60 and Fx cos 60 , but the answer is opposite.

9. Apr 16, 2013

### CWatters

Seems ok to me.

The two components are

F*Cos(60)
F*Sin(60)

Multiply by their distance from the pivot..

y*F*Cos(60)
x*F*Sin(60)

x*F*Sin(60) + y*F*Cos(60)

10. Apr 16, 2013

### cmkc109

it still seems to be Fx cos 60 + Fy sin 60 to me :/

11. Apr 16, 2013

### haruspex

You need to multiply the x by the component of F that's perpendicular to it, so that's the Y component, F sin(60o).
However, none of the offered answers is correct. The X and Y components of F act in opposite sense around A, so there should be a minus sign somewhere. E.g. consider the case where y = x tan(60o). There should be no moment.

12. Apr 17, 2013

### CWatters

Oops I missed that.

x*F*Sin(60) - y*F*Cos(60)

or

y*F*Cos(60) - x*F*Sin(60)

#### Attached Files:

• ###### Torque.jpg
File size:
12.5 KB
Views:
200
Last edited: Apr 17, 2013
13. Apr 17, 2013

### cmkc109

Thank you! makes sense now