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

Answer:

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

Step-by-step 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.

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

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

$F_{n} = 174.9 N$