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Harsh Torque problem. Check for me :o)

  1. Oct 27, 2005 #1
    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 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:




    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))

    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))

    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?

    Attached Files:

    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Oct 27, 2005 #2


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    Homework Helper

    I did not repeat the calculations in detail, but basically they seem correct, except that I think the moment arm of the quadriceps will stay 5 cm - see the attachment. Hold your lower leg at a 45o angle and feel the quadriceps muscle with your fingers - repeat it with the leg straight out - you should notice a difference in the hardness of the muscle. The reason why is that the moment arm is shortened when the limb is at an angle - the moment is larger when the distance to the object (think of the boot) is further away.

    Attached Files:

  4. Oct 27, 2005 #3
    So, wait wait, in the 180 position, the quad is still working upwards right? not horizontal because then it would be acting through the axis of rotation, which gives you no torque force.

    I get what you're saying, it says moment arm should be constant lol, my bad. But are you also saying that in the 180 position, the moment arm is larger and therefore that's why there is greater force required for the quad to maintain that position? But I thought the moment arm was constant?
  5. Oct 28, 2005 #4


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    The knee joint has quite a large diameter. If one assumes that the axis of rotation is in the middle of the knee joint, that is the contact point between the tibia and femur is in the middle of the joint, then the quadriceps tendon have a lever arm since it connects to the outer side of the tibia. Excuse my misuse of medical terminology, I did'nt even take biology in school. The quadriceps tendon pulls parallel to the axis of the tibia bone, whichever way the lower leg is pointing , well it is actually the patella ligament that pulls on the lower leg - the quadriceps tendon connects to the patella... which slips over the knee joint as the quadriceps pull on it. So you are right the lever stays constant, but in your calculation you multiplied it with the sine of the angle.
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