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

Question

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

Answer 1

unett

Answered 2021-04-23 Author has 119 answers

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Answer 2

Weight of the arm $F_{g} = 44.0 N$

Applying equlibrium conditions $\sum τ = 0$

$(F_{y}) (0.08 m) - (F_{g}) (0.290 m) = 0$

$(F_{t} \sin 12^{\circ}) (0.08 m) - (F_{g}) (0.290 m) = 0$

$F_{t} = \frac{(44.0 N) (0.290 m)}{(0.08) (\sin 12^{\circ})}$

$= 768.6 N$

$\approx 769 N$

Magnitude of tension force

$F_{t} = t = 769 N$

Now apply the first conditions of equlibtium

Along xaxis

$\sum F_{x} = 0$

$F_{x}^{'} - F_{t} \cos 12^{\circ} = 0$

$F_{x}^{'} = F_{t} \cos 12^{\circ}$

$F_{x}^{'} = (769 N) \cos 12^{\circ}$

$= 752.19 N$

Along y axis

$\sum F_{y} = 0$

$F_{t} \sin 12^{\circ} - F_{y}^{'} - 44.0 N = 0$

$F_{y} = - 44.0 N + F_{t} \sin 12^{\circ}$

$\approx 116 N$

So the force F, extered by the shoulder. On the humerces to hold the arms in the position

$F_{s} = s = \sqrt{F_{x}^{' 2} + F_{y}^{' 2}}$

$= \sqrt{(759.19 N)^{2} + (116 N)^{2}}$

$= 761 N$

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

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