Calculating Torque of Athlete's Arm at the Gym

[Jaklynn429]

Homework Statement

An athlete at the gym holds a 2.2 kg steel ball in his hand. His arm is 74 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?
1) Parallel out to his side
2) Straight but 35 M below the horizontal

Homework Equations

t=rf

The Attempt at a Solution

I don't even know how to start this one :(

 

[LowlyPion]

Homework Statement

An athlete at the gym holds a 2.2 kg steel ball in his hand. His arm is 74 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?
1) Parallel out to his side
2) Straight but 35 M below the horizontal

Homework Equations

t=rf

The Attempt at a Solution

I don't even know how to start this one :(

How does the distance of the ball and center of mass of the arm from his shoulder change horizontally? Isn't torque about a point the distance away through which the force acts? Hence the vertical force of gravity acts through the ball and center of mass straight down, so how far away along the horizontal in each of the 2 situations?

 

[Jaklynn429]

I got the first part by using
(4*.37) + (2.2*.74) = 3.108*9.8=30.46

Now I am having trouble with the second part, wouldn't I just use that answer times sin(35)? I get 17.468 and its telling me I am wrong.

 

[LowlyPion]

I got the first part by using
(4*.37) + (2.2*.74) = 3.108*9.8=30.46

Now I am having trouble with the second part, wouldn't I just use that answer times sin(35)? I get 17.468 and its telling me I am wrong.

Isn't Cos the horizontal distance?

 

[Jaklynn429]

I got the answer! thank you so much!