Calculating Quadriceps Force | Basic Torque Question Solution

[ally1h]

Homework Statement

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

Homework Equations

Torque = (Force)x(perpendicular distance)
Basic Trigonometry

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?

 

[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.

 

[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?

 

[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.

 

[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...

 

[rl.bhat]

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

 

[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.

 

[rl.bhat]

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.

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.

 

[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.

 

[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