# The arm of a crane at a construction site is 15.0 m long

1. Oct 19, 2011

### DrunkApple

1. The problem statement, all variables and given/known data
The arm of a crane at a construction site is
15.0 m long, and it makes an angle of 15.8
with the horizontal. Assume that the max-
imum load the crane can handle is limited
by the amount of torque the load produces
around the base of the arm.
What maximum torque can the crane with-
stand if the maximum load the crane can
handle is 779 N?
Answer in units of N · m

2. Relevant equations
torque = rFsin θ

3. The attempt at a solution
torque = radius * Force of gravity * sin θ
= 15 m * 779 N * sin 15.8
I don't get what is wrong...

2. Oct 19, 2011

### Delphi51

Re: Torque

"torque = 15 m * 779 N * sin 15.8"

Surely it should be cosine rather than sine?

3. Oct 19, 2011

### DrunkApple

Re: Torque

uhhh can i ask why it's cosine?

4. Oct 19, 2011

### Delphi51

Re: Torque

Torque is the vertical force times the horizontal distance from the pivot. The angle of the crane with horizontal is given. So 15*cos(15.8) is the horizontal component of the crane's length.