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

[DrunkApple]

Homework Statement

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

Homework Equations

torque = rFsin θ

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...

 

[Delphi51]

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

Surely it should be cosine rather than sine?

 

[DrunkApple]

uhhh can i ask why it's cosine?

 

[Delphi51]

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.

 

[asteriatic]

Your attempt at a solution is incorrect. The formula for torque is τ = rFsinθ, where r is the radius of the arm, F is the force applied, and θ is the angle between the arm and the force. In this case, the force is not the force of gravity, but rather the maximum load the crane can handle. So the correct formula would be τ = 15 m * 779 N * sin 15.8, which equals approximately 1,947 N · m. This is the maximum torque the crane can withstand with a maximum load of 779 N.