An arm held horizontallly is composed of three parts: Theupper arm with a weight

Question

An arm held horizontallly is composed of three parts: Theupper arm with a weight W1= 17N with lever arm 1=0.13m,the lower arm weight W2= 11N with lever arm 2= 0.38m,and the hand W3= 4.2N with lever arm =0.61m. Find thecenter of gravity of the entire arm relative to the shoulderjoint. a.   0.054m b.   0.14m c.   0.108m d.   0.28m What is the formula for this problem? An arm held horizontallly is composed of three parts: Theupper arm with a weight W1= 17N with lever arm 1=0.13m,the lower arm weight W2= 11N with lever arm 2= 0.38m,and the hand W3= 4.2N with lever arm =0.61m. Find thecenter of gravity of the entire arm relative to the shoulderjoint. a.   0.054m b.   0.14m c.   0.108m d.   0.28m What is the formula for this problem?

Explanation / Answer

In general the center of gravity along x-axis is the sum ofmoments xi Wi over the sum onforces Wi . X = xi Wi /Wi Then Then xi Wi   = 0.13 x 17+ 0.38 x 11 + 0.61 x 4.2= 8.95 and Wi = 17 + 11 + 4.2= 32.2N X= 8.95/ 32.2 X=0.28m The answer is d. The answer is d.

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