Shear force on the bone is the case where a force couple acts on the bone perpendicular to the lenght of the bone. A good example is when the leg is stuck against an abstacle and a force is applied from the opposite direction.
But in reality, there will never be zero torque at any given angle. There must have some torque in all angle given by the contraction of the quadriceps. Therefore, I think there is some mistakes?
When the position is static, (not moving), the sum of all the torques MUST equal zero. That is the fundamental principle of solving static torque problems.
I think I was confused when he said if the angle is 90deg there is no torque by the muscle, but now I realize that it will never be 90deg as the angle of the pulling force from patella tendon to the tibia will always have some degrees like around 25 degrees in anatomy.
I want to thank you for your work! I just have a couple of questions. First, why is the cos65 an easier angle than sin25? Second, does using cos65 form a new triangle? I have been struggling to picture it. Thanks again!
cos(65) is the same as sin(25), not easier. No, cos(65) does not form a new triangle. There are often multiple ways in which the same problem can be solved.
im really loving these videos on body mechanics! its so amazing to see, i didnt realize how strong are muscles really are
It is absolutely amazing.
You are truly a blessing Sir. Thanks so much for everything.! Physics is much more easier for me now.!
is their shear force when the legs aren’t fully extended or in any part of the movement
Shear force on the bone is the case where a force couple acts on the bone perpendicular to the lenght of the bone. A good example is when the leg is stuck against an abstacle and a force is applied from the opposite direction.
I dont understand how do you get cos 65. I understand that you substract, but in terms of lever arm? I need to see the triangle for F force?
The L/5 always becomes hypotenuse of a right triangle, and It is equal to use sin25 rather than cos65 (sin25 = cos65).
hwigeum Jeong thank u so much! All the best!
Nice, thanks for the info, well done :) so F = 8xW
Glad you liked it.
Thanks for the video. May you give us the literature that you used?
This is a homework problem I assigned a long time ago.
But in reality, there will never be zero torque at any given angle. There must have some torque in all angle given by the contraction of the quadriceps. Therefore, I think there is some mistakes?
When the position is static, (not moving), the sum of all the torques MUST equal zero. That is the fundamental principle of solving static torque problems.
I think I was confused when he said if the angle is 90deg there is no torque by the muscle, but now I realize that it will never be 90deg as the angle of the pulling force from patella tendon to the tibia will always have some degrees like around 25 degrees in anatomy.
Yes, that would be a correct statement.
Thank you so much!
You are welcome, glad you found our videos. 🙂
I want to thank you for your work! I just have a couple of questions. First, why is the cos65 an easier angle than sin25? Second, does using cos65 form a new triangle? I have been struggling to picture it. Thanks again!
cos(65) is the same as sin(25), not easier. No, cos(65) does not form a new triangle. There are often multiple ways in which the same problem can be solved.
who is confused with picture of problem?🤔🤔🤔🤔
Why did you delete my comment Sir? I just asked for any literatures titles including this problem.
Thanks...
Thnx
👍
I didnt got the point where you took cos65°
Torque (of F) = (L/5) x F x cos (angle with the vertical to the leg) angle = 90 - 25 = 65
Please translated to Arabia
A good example of how pointless university is im doing this for primary pe teaching
We don't always get what we pay for. Glad you found our videos.