Question

Three people pull simultaneously on a stubborn donkey. jack pulls directly ahead of the donkey with a force of 97.9 n, jill pulls with 72.7 n in a direction 45° to the left, and jane pulls in a direction 45° to the right with 145 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

We should analyze the forces in two different directions: the forward-backward direction and the left-right direction, and then calculate the resultant of the net forces acting on both directions.

Forward-backward direction. Here we have three forces acting on the monkey: the force applied by Jack, 97.9 N directly ahead; the force applied by Jill, 72.7 N with an angle
`45^(\circ)` to the left; and the force applied by Jane, 145 N with an angle
`45^(\circ)`. Therefore, the resultant on this axis is

`F_x = 97.9 N + 72.7 N\cdot \cos (45^(\circ)) + 145 N \cos (45^(\circ)) = 251.8 N`

Left-right direction. In this direction, the force applied by Jack is 0 N, because he is applying his force only ahead. The force applied by Jill is 72.7 N with an angle
`45^(\circ)` to the left, while the force applied by Jane is 145 N with an angle
`45^(\circ)` to the right: this means we should write the two forces with opposite signs, because they have opposite direction in the left-right axis. Therefore,

`F_y = 72.7 N \cdot \sin (45^(\circ))-145 N \cdot \sin (45^(\circ))=-51.1 N`

The net force acting on the monkey is the resultant of these two forces:

`F= √(F_x^2+F_y^2)= √((251.8N)^2+(-51.1N)^2)=257 N`