Question

1. a) A student hangs three masses from a force table. If the first mass of 250 g is at an angle of 130°, and she then hangs a mass of 205 g at 225°, calculate the mass and angle of the third mass so that the ring on the force table is in equilibrium? For maximum credit, show your work. Assume the given masses include the mass of the hangers b) the attached sheet of graph paper, represent the vectors from part a graphically and to scale. Each square on the graph paper is 25 g

Answer 1

the sum of the horizontal components are to be zero

0.25 X cos130 + 0.205 X cos 225 + m cos $\theta$ = 0

m cos $\theta$ = - 0.25 X cos130 - 0.205 X cos 225 = 0.306

and the sum of the vertical compnents is to be zero is

0.25 X sin130 + 0.205 X sin 225 + m cos $\theta$ = 0

m cos $\theta$ = - 0.25 X sin130 - 0.205 X sin 225 = - 0.0466

from horizontal and vertical components we get

tan $\theta$ = - 0.0466 / 0.306

tan $\theta$ = - 0.1523

$\theta$ = 351.339^o

m = 0.306 / cos351.339

m = 0.3095 kg

m = 309.5 grams

and

$\theta$ = 351.339^o