The total mass of the arm shown in (Figure 1: http://www. webassign. net/giancoli5/9-67alt. gif, where theta = 15 degrees) is 3. 5kg .

**Answer 1**

Sum the moments about the shoulder; only the vertical component of Fm and the weight of the arm contribute. The sum must be zero, or the arm would be rotating.

0 = 3.5kg * 9.8m/s² * 24cm – Fm * sin15º * 12cm = 823N·cm – 3.1cm * Fm

Fm = 265 N

which has components Fmv = 265N * sin15 = 68.6 N

and Fmh = 265N * cos15 = 256 N

So Fjh = 256 N ← so the horizontal forces sum to zero

and Fjv = 3.5kg * 9.8m/s² – 68.6N = -34.3 N (that is, down) ← so vertical forces sum to 0

A) Then Fj = √(Fjh² + Fjv²) = 258 N

B) Θ = arctan(34.3/256) = 7.6º