# Rotational problems

## Homework Statement

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?

## Answers and Replies

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Doc Al
Mentor
Use the cross product definition of torque:

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

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?

Doc Al
Mentor
ok so F = 1.5 i + 2.4 j
OK.
and x = -1.3 m, y = 2.4
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

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) ?

Doc Al
Mentor
so the r displacement would be 3-(-1.3) i + 0-2.4 j
Right.
so r would be sqrt( 4.3^2 + 2.4^2) ?
That would be the magnitude of r.

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

Doc Al
Mentor
ok so since i have 4.3 i - 2.4 j how do i begin to find the torque and such?
Take the cross-product, as I indicated in post #2.

so i get 13.92 k is that right?

Doc Al
Mentor
so i get 13.92 k is that right?
Sounds good to me. (Don't neglect the units.)

ok thanks a lot!

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