Map Sapling Teaming Three people pull simultaneously on a stubborn donkey. Jack

Question

Map Sapling Teaming Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 89.3 N. Jill pulls with 86.9 N in the northeast direction, and Jane pulls to the southeast with 129 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey Number What is the direction of the net force? Express this as the angle from the east direction between 0° and 90°, with a positive sign for north of east and a negative sign for south of east Number

Explanation / Answer

As given in the question,

F1 = 89.3 N, F2 = 86.9 N and F3 = 129 N

F1(x) = 89.3 N and F1(y) = 0 N

F2(x) = F2 cos45°  = 61.45 N and F2(y) = F2 sin45°  = 61.45 N

F3(x) = F3 cos45°  = 91.22 N and F3(y) = - F2 sin45°  = - 91.22 N

So for the net force,

FNET(x) = F1(x) + F2(x) + F3(x) = 241.97 N

FNET(y)  = F1(y) + F2(y) + F3(y) = - 29.77 N

So, the magnitude of net force,

FNET = sqrt [ {FNET(x)}^2 + {FNET(x)}^2} ]  = 243.79 N

And the required angle,

= tan^-1 [ FNET(y) / FNET(x) ] = tan^-1 [ (- 29.77) / (241.97) ] = - 7.014°

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