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.
3 answers
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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
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
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
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
1 answer