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The question is:
You are wearing a special 10 kg weight boot on each of your feet and you're sitting on a table with your legs hanging over the edge at the knee joint. The knees are held at an angle of 180 degrees (straight out) and 135 degrees. In each case how much force must be produced by each quadriceps muscle to hold the leg in this poistion? Assume that the patellar ligament has a constant moment arm of 5cm from the center of the knee joint. Assume the weight of the boot and foot are acting at the end of the leg.
Ok seems like a hard one, but I think it's alright, I hope :yuck: .
Listing all my relevant values:
Weight=68kg
Weight of your leg and foot=68kg x 6.18% (segment weights expressed as percentages of total body weight) = 4.2024 kg for each leg, the leg is just from the tibia and fibula to the foot not the thighs.
location of the center of mass of leg- so my leg part is 45 cm and the center of mass is supposed to be 43.4% of this length from the proximal patellar end, which comes to 19.53 cm.
Ok so attached is a drawing of my free body diagrams for both 180 and 135 degrees.
Also linked here:
http://www.geocities.com/mvxraven/leg180.JPG"
and
http://www.geocities.com/mvxraven/leg135.JPG"
Anyway, so I did the net torque routine:
for 180 degree: T net= sum of all(Fxd)
Remember F1 was the force of quadriceps (the unknown), F2 was the force due to center of mass of leg, and F3= force from 10kg boot
0=(F1xd1)+ (F2xd2)+(F3xd3)
0=(F1 x 0.05m) + ( (-9.81 x 4.02kg)) x 0.1953m) + ((-9.81x10kg) x 0.45m))
0=0.05F1-8.05-44.145
52.195/0005=F1
F1=1043.9 N. (Seems like a reasonable answer for something like holding a 10 kg boot up at 180 degrees)
Now for the 135:
0=(F1xd1)+ (F2xd2)+(F3xd3)
0=(F1 x 0.0389) + ( (-9.81 x 4.02kg)) x 0.135m) + ((-9.81x10kg) x 0.306m))
0=0.0389F1-5.323-30.018
35.3=0.0389F1
F1=908.5N (Ok this is strange, I thought you needed more force when the leg is at an angle vs when it is straight. Just like a leg extension exercise, when at the top, the weight is easier to hold than during the middle part)?
So PHEW! that took a while, I hope you guys can give me some input here...
are my diagrams correct, did my values make any sense?
You are wearing a special 10 kg weight boot on each of your feet and you're sitting on a table with your legs hanging over the edge at the knee joint. The knees are held at an angle of 180 degrees (straight out) and 135 degrees. In each case how much force must be produced by each quadriceps muscle to hold the leg in this poistion? Assume that the patellar ligament has a constant moment arm of 5cm from the center of the knee joint. Assume the weight of the boot and foot are acting at the end of the leg.
Ok seems like a hard one, but I think it's alright, I hope :yuck: .
Listing all my relevant values:
Weight=68kg
Weight of your leg and foot=68kg x 6.18% (segment weights expressed as percentages of total body weight) = 4.2024 kg for each leg, the leg is just from the tibia and fibula to the foot not the thighs.
location of the center of mass of leg- so my leg part is 45 cm and the center of mass is supposed to be 43.4% of this length from the proximal patellar end, which comes to 19.53 cm.
Ok so attached is a drawing of my free body diagrams for both 180 and 135 degrees.
Also linked here:
http://www.geocities.com/mvxraven/leg180.JPG"
and
http://www.geocities.com/mvxraven/leg135.JPG"
Anyway, so I did the net torque routine:
for 180 degree: T net= sum of all(Fxd)
Remember F1 was the force of quadriceps (the unknown), F2 was the force due to center of mass of leg, and F3= force from 10kg boot
0=(F1xd1)+ (F2xd2)+(F3xd3)
0=(F1 x 0.05m) + ( (-9.81 x 4.02kg)) x 0.1953m) + ((-9.81x10kg) x 0.45m))
0=0.05F1-8.05-44.145
52.195/0005=F1
F1=1043.9 N. (Seems like a reasonable answer for something like holding a 10 kg boot up at 180 degrees)
Now for the 135:
0=(F1xd1)+ (F2xd2)+(F3xd3)
0=(F1 x 0.0389) + ( (-9.81 x 4.02kg)) x 0.135m) + ((-9.81x10kg) x 0.306m))
0=0.0389F1-5.323-30.018
35.3=0.0389F1
F1=908.5N (Ok this is strange, I thought you needed more force when the leg is at an angle vs when it is straight. Just like a leg extension exercise, when at the top, the weight is easier to hold than during the middle part)?
So PHEW! that took a while, I hope you guys can give me some input here...
are my diagrams correct, did my values make any sense?
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