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The person weighs 170 lb. Each crutch makes an angle of 22.0 with the vertical. Half of the person's weight is supported by the cruches, the other hal

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asked 2021-03-17
The person weighs 170 lb. Each crutch makes an angle of 22.0 with the vertical. Half of the person's weight is supported by the cruches, the other half by the vertical forces exerted by the roundon his feet. Assuming that he is at rest and that the force exerted by the ground on the crutches acts along the crutches,determine
a) the smallest possible coefficient of friction between crutches and ground and
b) the magnitude of the compression force supported by each crutch.

Answers (1)

2021-03-19
image
image
free body diagram of cruches
from the first diagram (\(\displaystyle{n}_{{{g}{r}{o}{u}{n}{d}}}={\frac{{{170}}}{{{2}}}}={85}\ {l}{b}\))
\(\displaystyle\sum{F}_{{x}}={F}_{{1}}{\sin{{22}}}-{F}_{{2}}{\sin{{22}}}={0}\)
\(\displaystyle\sum{F}_{{y}}={2}{F}{\cos{{22}}}+{85}{l}{b}{s}-{170}{l}{b}{s}\)
so F=45.8lbs
From the second diagram
\(\displaystyle\sum{F}_{{x}}={f}-{45.8}{l}{b}{s}{\sin{{22}}}={0}\) f=17.2lbs
\(\displaystyle\sum{F}_{{y}}={n}_{{{t}{i}{p}}}-{45.8}{\cos{{22}}}={0}\) \(\displaystyle{n}_{{{t}{i}{p}}}={42.5}{l}{b}{s}\)
for minimum coefficient of friction
\(\displaystyle\mu={\frac{{{f}}}{{{n}_{{{t}{i}{p}}}}}}={0.404}\)
b) compressing force F =45.8lbs.
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