56 cnn 21 cnt (6 kg)g 42 kg)g Deltoid musc 6. The fellow pictured in the figure

Question

56 cnn 21 cnt (6 kg)g 42 kg)g Deltoid musc 6. The fellow pictured in the figure above has an arm of 4.2 kg mass, is 56 cm long and with a center of mass that is 21 cm from his shoulder. The hand at the end of the arm holds a 6.0 kg barbell a) What is the torque about the shoulder due to the weights of the arm and the 6.0 kg mass? b) If the arm is held in equilibrium by the deltoid muscle, whose force on the arm acts 5.0° below the horizontal at a point 18 cm from the shoulder joint (lower figure), what is the force exerted by the deltoid muscle. 4

Explanation / Answer

(A) torque = r x F

Net torque = (0.21 x 4.2 x 9.8 x cos15) + (0.56 x 6 x 9.8 x cos15)

= 40.2 N m . .....Ans

(B) torque due to deltoid muscle will be 40.2 N m.

40.2 = (0.18)(F) (cos(15-5))

F = 40.2/(0.18 x cos10)

F = 226.5 N

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.