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.

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.

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5