- #1
stunner5000pt
- 1,461
- 2
ok have a look at the diagram posted. When someone broke their femur (thigh bone) the muscles around the femur would compress tightly and thereby shorten the bone once it heals back. SO the apparatus described in the diagram is what they used to do to prevent the femur from growing back shorter. The question is to find the horizontal force - in terms of the mass,m and the mass of the leg(ML).
What i was thinking of doing was to assume that htis was a simply pulley system where there were two masses on either side of the one pulley... but that wouldn't yeild an answer since it would get (since there is no acceleration)
[tex] T_{1} - mg = T_{2} - m_{L}g [/tex]
but then i cnanot solve for T1 or T2 (can i?)
Now lookin at my diagram and my various labelled tensions...
T1 = mg
T2 T2cos 50 = mg and T2 = mg / cos 50
vertical component is t2 cos 50\
horizontal t2 sin 50
is this the same value for T3 ? When does trhe value of the other force (gravity on the leg) come into play?
Please help! Thank you very much!
What i was thinking of doing was to assume that htis was a simply pulley system where there were two masses on either side of the one pulley... but that wouldn't yeild an answer since it would get (since there is no acceleration)
[tex] T_{1} - mg = T_{2} - m_{L}g [/tex]
but then i cnanot solve for T1 or T2 (can i?)
Now lookin at my diagram and my various labelled tensions...
T1 = mg
T2 T2cos 50 = mg and T2 = mg / cos 50
vertical component is t2 cos 50\
horizontal t2 sin 50
is this the same value for T3 ? When does trhe value of the other force (gravity on the leg) come into play?
Please help! Thank you very much!
