Calculating Torque with Vectors at an Angle

[lukatwo]

Homework Statement

So I was wondering how do we account for vectors that are at a certain angle. The problem that I'm having with the picture is: if I was calculating torque from point B, how would I account for the vector G.

Homework Equations

The Attempt at a Solution

I've tried taking the component that's vertical to the slope, but in that case what is the lever arm? Is it where the vector G crosses the slope(point D)?

 

[SammyS]

Homework Statement

So I was wondering how do we account for vectors that are at a certain angle. The problem that I'm having with the picture is: if I was calculating torque from point B, how would I account for the vector G.

Homework Equations

The Attempt at a Solution

I've tried taking the component that's vertical to the slope, but in that case what is the lever arm? Is it where the vector G crosses the slope(point D)?

Try taking the component that's perpendicular to the ramp (incline).

Do you know about the line of action for a force -- in this case force G ?

 

[lukatwo]

I've tried taking the perpendicular component, but not sure what the lever arm is. I'm not sure how to determine the line of action.

 

[electronicsguy]

I've tried taking the perpendicular component, but not sure what the lever arm is. I'm not sure how to determine the line of action.

By extending the force T, you can see that it will contact point D. So it is like you're applying the force T directly on point D and the lever arm will be BD. Then take the perpendicular component of force T relative to the slope.

You can always check.
If we assume that BT is perpendicular to force T (note: BT is the dist. bet. B and T)
From the figure, the τ = T*(BT) . But, BT = BDsin(90-α). Therefore τ = T*(BDsin(90-α)) [1].
Returning to your problem, τ = Txr where r is the lever arm. The perpendicular component of T is Tsin(90-α). Therefore τ = Tsin(90-α)*r = T*(BDsin(90-α)) from [1]. Therefore r = BD.