# Rotational problems

1. Mar 13, 2009

### hellothere123

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

1) A force F = 1.5 i + 2.4 j N is applied at the point x = 3.0 m, y = 0 m. Find the torque about the point x = -1.3 m, y = 2.4 m.

I believe the lever arm should be the distance between the two points. but then the force that acts perpendicular to this is where i cannot get the right answer. How do i do this problem?

2. Mar 13, 2009

### Staff: Mentor

Use the cross product definition of torque:

$$\tau = \vec{r} \times \vec{F}$$

3. Mar 13, 2009

### hellothere123

ok so F = 1.5 i + 2.4 j and x = -1.3 m, y = 2.4
i get -3.6 K and -1.95K so is it.. -5.55 k?

4. Mar 14, 2009

### Staff: Mentor

OK.
No, that's just the point about which you are finding the torque. $\vec{r}$ is the displacement from that point: r = (x2 - x1) i + (y2 - y1) j

5. Mar 14, 2009

### hellothere123

F = 1.5 i + 2.4 j N is applied at the point x = 3.0 m, y = 0 m.
x = -1.3 m, y = 2.4

so the r displacement would be 3-(-1.3) i + 0-2.4 j
so r would be sqrt( 4.3^2 + 2.4^2) ?

6. Mar 14, 2009

### Staff: Mentor

Right.
That would be the magnitude of r.

7. Mar 14, 2009

### hellothere123

ok so since i have 4.3 i - 2.4 j how do i begin to find the torque and such?

8. Mar 14, 2009

### Staff: Mentor

Take the cross-product, as I indicated in post #2.

9. Mar 15, 2009

### hellothere123

so i get 13.92 k is that right?

10. Mar 15, 2009

### Staff: Mentor

Sounds good to me. (Don't neglect the units.)

11. Mar 15, 2009

### hellothere123

ok thanks a lot!

i ask if it was right because i am unable to check my answer.