Question

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 83.7 N , 83.7 N, Jill pulls with 73.1 N 73.1 N in the northeast direction, and Jane pulls to the southeast with 181 N . 181 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

Answer:

274.2 N

Explanation:

To find the magnitude of the net force, we have to resolve each force along two perpendicular directions and then calculate the components of the net force along these 2 directions.

The three forces are:

$F_1=83.7 N$ east, so positive x-axis

$F_2=73.1 N$ northeast, so at $\theta=45^{\circ}$ above positive x-axis

$F_3=181 N$ southeast, so at $\theta=-45^{\circ}$

The components of each force along the two directions are:

Force 1:

$F_{1x}=83.7 N\\F_{1y}=0$

Force 2:

$F_{2x}=(73.1)(cos 45^{\circ})=51.7N\\F_{2y}=(73.1)(sin 45^{\circ})=51.7 N$

Force 3:

$F_{3x}=(181)(cos (-45^{\circ}))=128.0 N\\F_{3y}=(181)(sin(-45^{\circ}))=-128.0 N$

So, the components of the net force along the two directions are:

$F_x=F_{1x}+F_{2x}+F_{3x}=83.7+51.7+128.0=263.4 N\\F_y=F_{1y}+F_{2y}+F_{3y}=0+51.7-128.0=-76.3 N$

Therefore, the magnitude of the net force is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(263.4)^2+(-76.3)^2}=274.2 N$