Homework Help: woburn

Posted by shariva on Monday, October 4, 2010 at 10:53pm.

We consider a man of mass m = 77 kg as shown in the figure below using crutches. The crutches each make an angle of è = 26 with the vertical. Half of the person's weight is supported by the crutches, the other half is supported by the normal forces acting on the soles of the feet. Assuming that the person is at rest, find the magnitude of the force supported by each crutch.

Physics needs help - PsyDAG, Tuesday, October 5, 2010 at 10:59am
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Physics - Gharib, Monday, October 25, 2010 at 12:28pm
m=77kg
angle=26 degrees
Normal force holds half the weight.

Positive being up,
Y direction:
(1/2)w-(F1)cos(26)-(F2)cos(26)
=(1/2)(77kg)(9.8m/s^2)-2Tcos(26) [since T1=T2
=> T=377.3N/2cos(26)
T=210N
Physics - Gharib, Monday, October 25, 2010 at 12:29pm
m=77kg
angle=26 degrees
Normal force holds half the weight.

Positive being up,
Y direction:
(1/2)w-(F1)cos(26)-(F2)cos(26)
=(1/2)(77kg)(9.8m/s^2)-2Fcos(26) [since F1=F2
=> F=377.3N/2cos(26)
F=210N

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