A) Determine the magnitude of the tension force F t in the deltoid muscle. B) De
ID: 1325268 • Letter: A
Question
A) Determine the magnitude of the tension force Ft in the deltoid muscle.
B) Determine the magnitude of the tension force Fs of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.
C) Determine the angle of tension force Fs relative to the x-axis.
The arm in the figure below weighs 43.5 N. The force of gravity acting on the arm acts through point A. Assume that L1 = 0.0890m, L2 = 0.315m and ? = 11.5o. A) Determine the magnitude of the tension force Ft in the deltoid muscle. B) Determine the magnitude of the tension force Fs of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown. C) Determine the angle of tension force Fs relative to the x-axis.Explanation / Answer
i have answered this question before....
The arm in the figure above weighs 37.7 N.
The force of gravity acting on the arm acts through point A. Assume that L1 = 0.0730m, L2 = 0.329m and a = 13.8 degrees
a.) Determine the magnitude of the tension force Ft in the deltoid muscle.
b.)Determine the magnitude of the tension force Fs of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.
c.) Determine the angle of tension force Fs relative to the x-axis.
ANS ) a) Vertical component of Ft = Ft Sin ?
Take moments about O:
Ft Sin ? L1 = Fg L2
Ft = 37.7 x 0.329 / 0.073 x Sin13.8
Ft = 12.4033 / 0.0174
Ft = 712.3 N
b) Horizontal component of Fs should be equal and opposite to horizontal component of Ft
Fs Cos ? = Ft Cos ? = 712.3 Cos 13.8 = 691.7 ............... (i)
And:
Taking moments about A:
Fs Sin ? = Ft Sin ?
Fs Sin ? = 712.3 Sin 13.8 = 169.91
From (i)
Fs Cos ? = 691.7
c) Fs Sin ? / Fs Cos ? = Tan ? = 169.91 / 691.7 = 0.2456
Angle ? = 13.8
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