Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead

Question

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 71.7 N, Jill pulls with 86.9 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 185 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 straight ahead between 0° and 90° with a positive sign for angles to the left and a negative sign for angles to the right. Number

Explanation / Answer

Net force will be given by:

Fnet = F1 + F2 + F3

F1 = 71.7 N in +ve x-axis

F1x = 71.7*cos 0 deg = 71.7 N

F1y = 71.7*sin 0 deg = 0 N

F2 = 86.9 N in 45 deg Clockwise to +ve x-axis

F2x = 86.9*cos 45 deg = 61.45 N

F2y = -86.9*sin 45 deg = -61.45 N

F3 = 185 N in 45 deg Counter-Clockwise to +ve x-axis

F3x = 185*cos 45 deg = 130.81 N

F3y = 185*sin 45 deg = 130.81 N

Net force will be

Fnet = Fnet)x + Fnet)y

Fnet = (71.7 + 61.45 + 130.81) i + (0 - 61.45 + 130.81) j

Fnet = 263.96 i + 69.36 j

|Fnet| = sqrt (263.96^2 + 69.36^2)

|Fnet| = 272.92 N

Direction will be given by:

theta = arccos [(Fnet)y/(Fnet)x]

theta = arccos (69.36/263.96) = 74.76 deg to the right

theta = +74.76 deg

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