Question

E,: 528 grams at 270 Now, please hang these equilibrating masses. Do they bring the ring into equilibrium? If not,what (34) went wrong. This state of affairs is represented below in Figure Eig Figure Eight 0or 67.8 247.8° 30° E,: 215 grams 528 grams One for You to Calculate 13,) Clear the force table of all hanging masses. Hang a total mass of 350 grams at the 25e mark, a total mass of 400 grams at the 75 mark, and a total mass of 300 grams at the 1250 mark. Calculate the magnitude of the t and y components of the force needed to 25e mar k, a total mass of 400 grams at the 75° mark, and a total mass of 300 grams at the equilibrate these three weights. grams at (35) (36) grams at Now, please hang the equilibrating masses. Do they bring the ring into equilibrium? If not, what went wrong On the graph provided below in Figure Nine, draw a correct, scaled representation of this state of affairs-see Figure Eight for a model. HI 2-7

Answer 1

Let F be equilibrium force. F bar + F_1 bar + F_2 bar + F_3 bar = 0. F bar + 350 (cos 25 degree i bar + sin 25 degree j bar) + 300 [(cos 125 degree) i bar + sin 125 j bar)] + 400 [cos 75 degree i bar + sin 75 j bar] = 0. F bar + 350 [0.906 i bar + 0.422 j bar) + 300 [-0.573 i bar + 0.819 j bar] + 400 (0.2588 i bar + 0.9659 j bar] = 0 F bar + 428.72 i bar + 779.79 j bar = 0 F bar = -248.72 i bar + 779.79 j bar N vert F bar vert = 8181.495