Simple torque equation question

[auk411]

Homework Statement

From the picture, take the moment about B. Find an equation for F_a

Homework Equations

rFsin(theta)

The Attempt at a Solution

Let r be some arbitrary distance.

rTsin(theta)-F(y) = Torque

Correct answer (I believe) is F_a = Tcos(theta)/ (y/L)

I have no idea how they got this. I thought Tcos(theta) went through the pivot point and hence had a torque of zero.

 

[PhanthomJay]

It does, but what about the torque from Tsin theta?? The distance 'r', depending on what you are defining as 'r', is not arbitrary. And the sum of the 2 torques must equal_____?

 

[auk411]

It does, but what about the torque from Tsin theta?? The distance 'r', depending on what you are defining as 'r', is not arbitrary. And the sum of the 2 torques must equal_____?

It must equal zero.

But to get the answer F_a = LTcos(theta).

But F_a isn't zero. What am I missing.

 

[PhanthomJay]

I believe you may have a misunderstanding as to how to calculate Torque.
There are 2 ways to calculate the torque (or also called moment) of a force or force component about a pivot point.

The first way is to calculate moment (M) is to break up the force into its x and y components, then sum the components of each Force (F) times the perpendicular distance (d_{perpendicular} ) from the line of action the force to the pivot point.
(M = sum of F(d_{perpendicular}), and watch plus and minus signs.

The 2nd way is M = r X F = (r)(F)(sin alpha), where r is the distance from the point of application of the force to the pivot point, and alpha is the angle in between F and r at the application point.
Note that the angle alpha in this equation is NOT thr angle theta given in the problem.

Use clockwise moments as minus, and counterclockwise moments as plus.

 

[auk411]

Yes, so how does this help me?

 

[PhanthomJay]

Yes, so how does this help me?

I trust you are not asking me to give the answer, but here are some tips:
You can use either approach.
Using the force time perpendicular method, then the moment about B is F_a(y) - (Tsin theta*(distance measured along the line AB). You'll have to do a bit of trig to calculate this distance. Note that the direction of Tsintheta and Tcostheta as shown in your figure, at point A, are, unfortunately, incorrect; they should point down and left, respectively.

Alternatively, using the r X F method, then the moment about B is F_a(y)(sin 90) - T(L) sin alpha, where alpha, as I defined earlier, is determined using the geometry of a right triangle.

 

[auk411]

Yes, but I don't get F_a = Tcos(theta)/ (y/L)

 

[PhanthomJay]

Please show your work and I can then see where your error might be.