Question

A wrench 30 cm long lies along the positive y-axis and grips bolt at the origin. A force is applied in the direction <0,3,-4> at the end of the wrench. Find the magnitude of the force needed to supply 100 N*m of torque to the bolt.

Answer 1

To answer the question, you need to have the force dispersed at (0.6) in the y-direction, and (0.8) in the z-direction (0.6² + 0.8² = 1) to supply 100 N*m of torque to the bolt. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Answer 2

= r x F
| τ | = | r x F | = | r | | F | sinΘ
100 = | (0.3)<0, 1, 0> x (F/5)<0, 3, -4> |
5000/3 = | F | | <0, 1, 0> x <0, 3, -4> |
5000/3 = | F | | <-4, 0, 0> |
5000/3 = | F | * 4
| F | = 1250/3
| F | = 417 N

Check:
τ = r x F
100<-1, 0, 0> = 0.3<0, 1, 0> x 250/3<0, 3, -4>
100<-1, 0, 0> = 25 ( <0, 1, 0> x <0, 3, -4> )
100<-1, 0, 0> = 25 ( <-4, 0, 0> )
100<-1, 0, 0> = 100<-1, 0, 0>

What the other people forgot was that you need the angle from the y-axis, which would be Θ = arctan(O/A) = arctan(z/y) = arctan(-4/3) = -53.1°. Then you take the sine of that to get 0.8. There is more work being done on the z-vector than there is on the y-vector, so the force is distributed (0.6) in the y-direction, and (0.8) in the z-direction (0.6² + 0.8² = 1).