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2021-04-21

The arm weighs 44.0 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force t in the deltoid muscle and the force s of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.

t=N

s=N

t=N

s=N

unett

Skilled2021-04-23Added 119 answers

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Jeffrey Jordon

Expert2021-09-29Added 2605 answers

Weight of the arm ${F}_{g}=44.0\text{}N$

$\sum \tau =0$

$({F}_{y})(0.08\text{}m)-({F}_{g})(0.290\text{}m)=0$

$({F}_{t}\mathrm{sin}{12}^{\circ})(0.08\text{}m)-({F}_{g})(0.290\text{}m)=0$

${F}_{t}=\frac{(44.0\text{}N)(0.290\text{}m)}{(0.08)(\mathrm{sin}{12}^{\circ})}$$=768.6\text{}N$$\approx 769\text{}N$

${F}_{t}=t=769\text{}N$

$\sum {F}_{x}=0$

${F}_{x}^{\prime}-{F}_{t}\mathrm{cos}{12}^{\circ}=0$

${F}_{x}^{\prime}={F}_{t}\mathrm{cos}{12}^{\circ}$

${F}_{x}^{\prime}=(769\text{}N)\mathrm{cos}{12}^{\circ}$

$=752.19\text{}N$

$\sum {F}_{y}=0$

${F}_{t}\mathrm{sin}{12}^{\circ}-{F}_{y}^{\prime}-44.0\text{}N=0$

${F}_{y}=-44.0\text{}N+{F}_{t}\mathrm{sin}{12}^{\circ}$$\approx 116\text{}N$

${F}_{s}=s=\sqrt{{F}_{x}^{\prime 2}+{F}_{y}^{\prime 2}}$$=\sqrt{(759.19\text{}N{)}^{2}+(116\text{}N{)}^{2}}$$=761\text{}N$

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