A woman lifts a 3.6-kg barbell in each hand with her arm in a horizontal positio

Question

A woman lifts a 3.6-kg barbell in each hand with her arm in a horizontal position at the side of her body and holds it there for 3 s (see the figure below). (Figure 1) What force does the deltoid muscle in her shoulder exert on the humerus bone while holding the barbell? The deltoid attaches 13 cm from the shoulder joint and makes a 13 angle with the humerus. The barbell in her hand is 0.55 m from the shoulder joint, and the center of mass of her 4.0-kg arm is 0.24 m from the joint. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

The total weight of the arm and the object in the hand are supported by two forces

The two forces are the vartical component of the tension in the deltoid musicle and up ward force which the scapula exerts on the sholder joint

T(arm) =M*g*x =4*9.81*0.24 =9.4176N

T(weight)=M*g*x=3.6*9.81*0.55=19.4238N

T(delta) =F*x=sin(13)*13m

Now balance the forces

T(delta) =T(weight)+T(arm)

T*0.13sin(13) =9.416N+19.4238N

T =986N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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