Question

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 95.5 N
, Jill pulls with 77.9 N
in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 133 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.

Answer 1

Let direction ahead of the donkey coincides with direction of the x-axis, and direction to the left of the donkey coincides with the y-axis.

Then forces applied to the donkey are:

$(95.5,0),\\ (77.9cos(45),77.9sin(45))=(77.9/\sqrt{2},77.9/\sqrt{2})\\ (133cos(-45),133sin(-45))=(133/\sqrt{2},-133/\sqrt{2})$ ,

the net force the people exert on the donkey

$F=(95.5,0)+(77.9/\sqrt{2},77.9/\sqrt{2})+(133/\sqrt{2},-133/\sqrt{2})=\\ (95.5+77.9/\sqrt{2}+133/\sqrt{2},0+77.9/\sqrt{2}-133/\sqrt{2})$

Magnitude of F is

$|F|=\sqrt{(95.5+77.9/\sqrt{2}+133/\sqrt{2})^2+(0+77.9/\sqrt{2}-133/\sqrt{2})^2}\approx$

$247.712N.$

Answer: the magnitude of the net force the people exert on the donkey.is $247.712N.$