Question

Three masses are hung on a force table. They are 60 g hung at 20 °, 20 g hung at 88 °, and 45 g hung at 152 °. Find where a fourth mass must be hung and how much the mass has to be to balance the ring in the center of the force table. Show your work in the space below. 

Answer 1

$$F_x=9.8(0.06\cos{20}+0.02\cos{88}+0.045\cos{152}) \\=0.17\ N$$

$$F_y=9.8(0.06\sin{20}+0.02\sin{88}+0.045\sin{152}) \\=0.604\ N$$

$$m(9.8)=\sqrt{0.17^2+0.604^2}\\m=64\ g$$

$$\theta=270-\arctan{\frac{0.17}{0.604}}=254\degree$$