Homework 4 – Problem 4
Monday, February 15th, 2016
4. The weightlifter was in the process of starting the second phase of a clean-and-jerk maneuver when his patellar tendon ruptured. The weight lifter weighs 980 N. The lower legs and feet account for 13% of his total weight when combined. The length of the lower leg is 40 cm. The knee can support both forces and moments. The mass of the load is 200 Kg.
THERE ARE THREE (3) PARTS TO THIS QUESTION:
4a.What is the total joint reaction at the knee when the weightlifter’s lower leg is at an angle of 70 degrees with the true horizontal as shown in Figure 3 above?
4b. What is the total joint reaction at the knee when the weightlifter’s lower leg is at an angle of 60 degrees with the true horizontal?
4c. Estimate the force in the patellar tendon in both cases.
422 thoughts on “Homework 4 – Problem 4”
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Anyone know the distance to the patellar tendon attachment, and how to determine the angle it’s force is at?
Does anyone have any ideas what we should use for the angle that the patellar tendon makes with the lower leg? I was thinking something small ~10deg…
Erik your comment popped up just after I submitted, ya I am wondering the same thing…
I was using 10 degrees; I have that number in my notes from lecture
Perfect! I did some rough measurements on my knee and calculated the angle and came up with 14 degrees.
Which lecture date was this from?
It is from lecture 14 (2/12/2016). He stated it when talking about the leg extension machine FBD.
In one of the professor’s lectures he said the angle was 10 degrees and does not change with respect to the position of the knee very much. I used 10 degrees. As far as the moment arm the tendon creates i searched the internet and found values ranging from 4cm to 6 cm i just used 5 as a average.
The question says “what are the reactions at the knee assume that the knee can support both forces and moments.” To me, this is instructing to do a total joint reaction evaluation, not find the Force of the Patella Tendon.
Did anyone else read the problem that way?
Yes, I understood it the same way. There’s a little bit more direction in the instructions above that confirms it as well.
I took the problem to use the total joint reaction model for parts a and b and the single force no moment model using the patellar tendon force for part c for both cases.
Agree, I am going to and use the knee as the point about which I take moments from for part c as well right?
We need the following information or to assume it:
frictional force between floor and shoe.
Center of pressure of the normal force from the floor onto the shoe (the location of the normal force)
If I assume the center of pressure is directly below the knee then the force of the PT is negative in this model which is not true.
I notice I can activate my quad and put tension in my PT, but it also activates other muscles.
Does anyone have a good justification for these assumptions?
I would assume the center of contact for the normal force to be very close to the heel as this is proper technique. Also, I assumed the horizontal friction force from the floor to be zero.
For the last part, if your leg is already static, then activating your quad to put tension on the pt requires the hamstrings to also fire to balance out the force and keep your leg from moving.
thanks. I’ve not performed a clean and jerk. I’ll incorporate the center of pressure at the heal. This makes for a positive PT load.
Generally speaking, in weightlifting the goal is to keep your weight on the center of your foot (it crosses the centerline of your body for balance purposes). In the jerk, this holds true until you actually do the jerk at which point the pressure goes to the toes for a moment then returns to the center foot.
“It’s important we drive through the heels during this step.”
http://breakingmuscle.com/olympic-weightlifting/a-step-by-step-guide-to-a-perfect-jerk
Could we use SFNM model for the entire problem? I am not exactly sure how to use the 70 vs 60 degree information, I’m thinking it would change the position of the force of the floor on the foot?
I used the 70 and 60 degree information to estimate the moment arms that the weight of the lower leg and floor force act through to result in a moment about the knee joint.
Are we supposed to find the forces in only one knee/patellar tendon or the total forces in both knees? To me, it makes more sense to do the force analysis on one knee and find a single patellar tendon force (I assumed that the body and bar weight were equally distributed on each foot). I guess if you found the total forces in both knees you could just divide by two (with the same weight distribution assumption).
How can we incorporate the 13% of his body weight into the FBD? Does this weight act downward in the middle of his leg?
put the 0.13*980N halfway down the lower leg, 20 cm from the knee at the angle given. You know that the vertical force or normal force from the floor onto the foot is equal to the total weight of the lifter plus the weight of the 200 kg bar. Thus your FBD and vertical static force summation leave as the unknown the force at the knee which is ok because you will sum moments about this point.
That is exactly what I did.