Finding the force of the femur on the knee cap.

  • Thread starter Thread starter meaganl
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The discussion focuses on calculating the force exerted by the femur on the kneecap, given a force of 420 N applied by the quadriceps muscle through a tendon. The model simplifies the leg's anatomy, treating the femur and tibia as beams connected by a pin, with the kneecap acting as a pulley. The user attempts to resolve the forces in both x and y components, resulting in a calculated femur force of approximately 191.4 N in the x-direction and -1.67 N in the y-direction. There is also a request for a larger diagram to better visualize the problem. The conversation highlights the complexities of biomechanics in force analysis.
meaganl
Messages
4
Reaction score
0

Homework Statement



The diagram of the leg shows the femur (1) and tibia (2). The quadriceps muscle (3) applies a force to the lower leg via a tendon (4) that is embedded with the kneecap (5). If the force applied by the muscle to the tendon is FM = 420 N, what is the force of the femur on the kneecap? A simplified model of the leg is shown next to the diagram. The leg bones are represented by two beams attached by a pin. The tendon is modeled by a rope and the kneecap acts like a pulley. The tendon above the kneecap makes an angle θ1 = 23° with respect to the vertical, and the portion of the tendon below the kneecap makes an angle of θ2 = 24° with respect to the vertical. Enter the x component, followed by the y component.

https://www.physicsforums.com/attachment.php?attachmentid=51494&thumb=1&d=1349314642

Homework Equations



fnet=ma



The Attempt at a Solution



I tried:

Fnetx=max
T1x + T2x + Ffemurx = 0
-240sin23 + -240sin24 + Ffemurx= 0
-93.78 -97.62 + Ffemurx=0
Ffemurx= 191.4N

Fnety=may
T1y + T2y + Ffemury=0
240cos23 -240cos24 + Ffemury=0
220.92-219.25 + Ffemury=0
Ffemury=-1.67N
 
Physics news on Phys.org
I see only very small picture. Can not you show a bigger one? ehild
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Calculating force of the femur on the knee cap.
Replies
1
Views
6K

Hot Threads

Recent Insights

Back
Top