# Basic torque question

1. Sep 14, 2009

### ally1h

1. The problem statement, all variables and given/known data
Find the Quadriceps force (Q) necessary to maintain equilibrium in this system.
G= 5#; G distance= 8"
W= 10#; W distance= 15"
Q=? ; Q distance= 1.5"
Theta 1 = 30 degrees
Theta 2 = 15 degrees

Here is a picture of the diagram:
http://farm3.static.flickr.com/2430/3920771451_38b594def2.jpg

2. Relevant equations
Torque = (Force)x(perpendicular distance)
Basic Trigonometry

3. The attempt at a solution
Torque of G= (5#)(8") = 40"#
Torque of W= (10#)(15")= 150"#
Sum of Torque = 190"#

What I can't remember for the life of me is how to get the torque for Q. I draw a perpendicular line... but I can't remember how to get the torque without knowing the force. After that, I should be fine with figuring out the rest of the problem. Please help?

2. Sep 14, 2009

### rl.bhat

Resolve Q into vertical and horizontal components.Torque due to G, W and the horizontal component of Q will be clockwise.
Torque due to vertical component of Q will be counterclockwise. For equilibrium condition, equate them.
While taking torque you have to the perpendicular distance from the pivot, not the actual distance.

3. Sep 14, 2009

### ally1h

So for the horizontal component:
[sin(15)] x [1.5"] = 0.39

And the vertical component:
[cos(15)] x [1.5"] = 1.45

For the sum of torques:
(40"#) + (150"#) + (0.39) = 190.39"#
190.39"# - 1.45 = 188.94"#

So for equilibrium, Q has to be 188.94?

Yes? No?

4. Sep 14, 2009

### rl.bhat

No.
If A is the point of application of Q,what is the angle between Q and AX and Q and AY?
The torque due to 5# and 10# are 5*8*cosθ2, and 10*15*cosθ2.
Similarly find the torque due to Q.

5. Sep 14, 2009

### ally1h

I'm sorry... I don't know. I was never good at Trig or Geometry and I haven't had to do any for the last 8 years. Is 30 too obvious an answer? I really have no clue...

6. Sep 14, 2009

### rl.bhat

Taking the hints from post 4#, show your calculations.

7. Sep 14, 2009

### ally1h

Tg = [cos(15)] x 5# x 8" = 38.64"#
Tw = [cos(15)] x 10# x 15" = 144.89"#

Qx = cos(15) x 1.5 = 1.45
Qy = sin(15) x 1.5 = 3.89

I must still be getting something wrong. From point A, the angle <QAX is 15 degrees and the angle <QAY is 90 degrees.

8. Sep 14, 2009

### rl.bhat

Qx and Qy are unknown quantities.
You have not used θ1 = 30 degrees any where.
From the figure find the angle between Q and vertical and Q and horizontal.

9. Sep 14, 2009

### ally1h

Listen, thank you for your time and effort, but your hints aren't helping me. I'm getting beyond frustrated because of my inability to do such a simple problem. I can't figure this out. I need someone to show me HOW to do this problem, not just give me hints and confuse me further, because that is all this is doing.

As I said, thank you for your time and effort, but if you can't show me HOW to do this problem I'm going to have to show up with incomplete homework and hope that my professor will explain it to me.

10. Sep 14, 2009

### rl.bhat

Angle θ1 is 30 degrees and θ2 = 15 degrees. Simple geometry shows that tha angle between vertical and Q is 45 degrees. Hence angle between horizontal and Q is also 45 degrees.
Now Clockwise torque = G*8*cos15 + W*15*coa15 + Q*cos45*1.5*sin15
Counterclockwise torque = Q*cos45*1.5*cos15.
I am not able to draw the diagram.
Equate the two torques and solve for Q