Question

three masses are hung on a force table. they are 60 grams hung at 20 degrees, 20 grams hung at 88 degrees, and 45 grams hung at 152 degrees. find where a forth mass must be hung and how much the mass has to be to. balance the ring in the center of the force table

Answer 1

Let's first find the $x$- and $y$-components of the resultant force:

$$\sum F_x=F_{1x}+F_{2x}+F_{3x},$$ $$60\ g\cdot cos20^{\circ}+20\ g\cdot cos88^{\circ}+45\ g\cdot cos152^{\circ}=17.35\ g,$$ $$\sum F_y=F_{1y}+F_{2y}+F_{3y},$$ $$60\ g\cdot sin20^{\circ}+20\ g\cdot sin88^{\circ}+45\ g\cdot sin152^{\circ}=61.63\ g.$$

We can find the magnitude of the resultant force from the Pythagorean theorem:

$$F=\sqrt{F_x^2+F_y^2}=\sqrt{(17.35\ g)^2+(61.63\ g)^2}=64\ g.$$

We can find the direction from the geometry:

$$\theta=cos^{-1}(\dfrac{F_x}{F})=cos^{-1}(\dfrac{17.35\ g}{64\ g})=74.3^{\circ}.$$

To find the angle on the force table we must subtruct $74.3^{\circ}$ from $360^{\circ}$:

$$\theta=360^{\circ}-74.3^{\circ}=285.7^{\circ}.$$

Therefore, the 64 g mass at 285.7 degrees will balance the ring in the center of the force table.