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 e force of 83.9 N, Jill pulls with 67.1 N In the northeast direction, and Jane pulls to the southeast with 137 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

Let the force exerted by Jack = F1

Force exerted by Jill = F2

Force exerted by Jane = F3

Let North be +y , then east is +x, south is -y and west is -x axis

F1 = 83.9 N in east direction

F1 = 83.9 i N

F2 = 67.1 in North-East direction (or, 45 deg anticlockwise from + x axis, thus angle = 45 deg)

F2 = 67.1 cos 45 i + 67.1 sin 45 j N

F3 = 137 South-East direction (or, 45 deg clockwise from + x axis, thus angle is -45 deg)

F3 = 137 cos (-45) i + 137 sin(-45) j N

Net Force:

F = F1 + F2 + F3

= 83.9 i + 67.1 cos 45 i + 67.1 sin 45 j + 137 cos (-45) i + 137 sin(-45) j

F = 228.22 i - 49.42 j

Magnitude of net force:

F = square root (228.222 + (-49.42)2) = 233.51 N

Angle from the east direction:

theta = tan-1 (-228.22/49.42) = -77.8 degrees

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