Torque about P (Cross Product?)

Reefy
Messages
62
Reaction score
1

Homework Statement



2mnn6l2.png



Homework Equations



The book says that

Torque = the length of the moment arm X the magnitude of the vertical component of F
= r X F

The Attempt at a Solution



My problem is that I don't have any vectors to do a cross product. Instead I tried to just multiply what I know:

50lbs x 9in x sin60° ≈ 389.71ft-lb

which is far from the correct answer, 32.5ft-lb. The problem seemed simple enough so am I making some small mistake or going about this completely wrong?
 
Physics news on Phys.org
Just because you aren't given any explicit vectors doesn't mean you can't formulate your own.

Looking at the problem, suppose that we set up an x-axis along the wrench.
In this case, vector r is (9,0) inches, with the emphasis on inches

The force vector F is going to be (50 cos (-60), 50 sin (-60)) lbs.

You can then evaluate r x F in the usual manner

Remember: in x lbs is not equal to ft-lbs
 
Ah, you're right! Thank you, that makes perfect sense.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top