Question

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 90.5 N , Jill pulls with 82.3 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 125 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

Fn= 174.9 N : Magnitude of the net force the people exert on the donkey.

Explanation:

We find the components of the forces in x-y-z

Force of Jack in z =F₁z=90.5 N in direction (+z)

Force of Jill in x = F₂x= -82.3*cos45°= - 58.19 N (-x)

Force of Jill in y =F₂y=-82.3*sin45°= + 58.19 N (+y)

Force of Jane in x =F₃x=125*cos45°= + 88.4 N (+x)

Force of Jane in y =F₃y=125*sin45°= + 88.4 N (+y)

Calculating of the components of the net force the people exert on the donkey.

Fnx= F₂x+F₃x=( - 58.19+ 88.4 )N=30.2N (+x)

Fny= F₂y+F₃y=( 58.19+88.4 ) = 146.59 N (+y)

Fnz =F₁z=90.5 N (+z)

Calculating of the magnitude of the net force the people exert on the donkey.

`F_(n) =\sqrt{(F_(nx))^(2)+(F_(ny)) ^(2) +(F_(nz)) ^(2)   }`

`F_(n) =\sqrt{(30.2)^(2)+( 146.59) ^(2) +(90.5) ^(2)   }`

`F_(n) = 174.9 N`