Question

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 76.9 N, Jill pulls with 70.9 N in the northeast direction, and Jane pulls to the southeast with 163 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

Fnet = F1+ F2 + F3

F1 = 76.9 N east

F2 = 70.9 N 45 deg N of E

F3 = 163 N 45 deg S of E

then, we know that

Suppose given that Vector is R and it makes angle A with +x-axis, then it's components are given by:

Rx = R*cos A

Ry = R*sin A

Using above rule:

F1 = 76.9 N & angle = 0 deg with +ve x-axis

F1x = F1*cos A1 = 76.9*cos 0 deg = 76.9 N

F1y = F1*sin A1 = 76.9*sin 0 deg = 0

F2 = 70.9 N & angle = 45 deg N of E

F2x = F2*cos A2 = 70.9*cos 45 deg = 50.13 N

F2y = F2*sin A2 = 70.9*sin 45 deg = 50.13 N

F3 = 163 N & angle = 45 deg below +ve x-axis = 315 deg (4th quadrant)

F3x = F3*cos A3 = 163*cos 315 deg = 115.26 N

F3y = F3*sin A3 = 163*sin 315 deg = -115.26 N

Now

Fnet = Fnet)x + Fnet)y

Fnet = (F1x + F2x + F3x) i + (F1y + F2y + F3y) j

Using above values

Fnet = (76.9 + 50.13 + 115.26) i + (0 + 50.13 - 115.26) j

Fnet = 242.29 i - 65.13 j

Direction of Fnet will be

Direction = arctan (Fnet_y/Fnet_x)

Direction = arctan (-65.13/242.29) = -15 deg = 15 deg below +ve x-axis

|Fnet| = magnitude of net force = sqrt (Fnet_x^2 + Fnet_y^2)

|Fnet| = sqrt ((242.29)^2 + (-65.13)^2)

|Fnet| = 250.89 N

Comment below if you have any doubt.