= 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).
