Answer to Question #176698 in Physics for Alysa Haden

Question #176698

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


1
Expert's answer
2021-03-31T07:14:09-0400

Let's first find the xx- and yy-components of the resultant force:


Fx=F1x+F2x+F3x,\sum F_x=F_{1x}+F_{2x}+F_{3x},60 gcos20+20 gcos88+45 gcos152=17.35 g,60\ g\cdot cos20^{\circ}+20\ g\cdot cos88^{\circ}+45\ g\cdot cos152^{\circ}=17.35\ g,Fy=F1y+F2y+F3y,\sum F_y=F_{1y}+F_{2y}+F_{3y},60 gsin20+20 gsin88+45 gsin152=61.63 g.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=Fx2+Fy2=(17.35 g)2+(61.63 g)2=64 g.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:


θ=cos1(FxF)=cos1(17.35 g64 g)=74.3.\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 74.374.3^{\circ} from 360360^{\circ}:


θ=36074.3=285.7.\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.


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