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 96.9 N, Jill pulls with 76.9 N in the northeast direction, and Jane pulls to the southeast with 131 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 0° and 90°, with a positive sign for north of east and a negative sign for south of east.

Explanation / Answer

here

force by jack F1 = 96.9 i

force by jill F2 = 76.9 / sqrt2 i + 76.9 / sqrt2 j

force by jane = 131 / sqrt(2) i - 131 / sqrt(2) j

net force F

F = F1 + F2 + F3

F = 96.9 i + (76.9 / sqrt(2) i + 76.9 / sqrt(2) j ) + (131 / sqrt(2) i - 131 / sqrt(2) j )

F = 243.9 i - 38.2 j

magnitude of the net force is

sqrt(243.9^2 + 38.2^2) = 246.87 N

then

theta = tan^-1(38.2 / 243.9 ) = 8.901 deg

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