Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with

Question

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 79.5 N, Jill pulls with 67.3 N in the northeast direction, and Jane pulls to the southeast with 155 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. What is the direction of the net force? Express this as the angle from the east direction between 0degree and 90degree, with a positive sign for north of east and a negative sign for south of east.

Explanation / Answer

Here,

force of Jack , F1 = 79.5 i N

force of Jill , F2 = 67.3 *(cos(45) i + j * sin(45)) N

FOrce of Jane , F3 = 155 *(cos(45) i - j * sin(45)) N

for the net force of donkey

Fnet = F1 + F2 + F3

Fnet = 79.5 i + 67.3 *(cos(45) i + j * sin(45)) + 155 *(cos(45) i - j * sin(45))

Fnet = 236.7 i - 62.01 j

Fnet = 244.7 N

the magnitude of net force is 244.7 N

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direction of net force = - arctan(62.01/236.7)

direction of net force = -14.7 degree

the direction of net force is -14.7 degree

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