Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 68.5 N, Jill pulls with 61.3 N in the northeast direction, and Jane pulls to the southeast with 111 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.What is the direction of the net force? Express this as the angle from the east direction between 0° and 90°, with a positive sign for north of east and a negative sign for south of east.

This is a routine vector addition problem.

Separately add the x (east) and y (north) components of the three force vectors.

Fjack,x= 90.7
Fjack,y = 0
Fjill,x= 89.2 cos45 = 63.07
Fjill.y = 89.2 sin45 = 63.07
Fjane,x = 175cos45 = ______
Fjane,y = -175cos45 = ______

For the resultant,
Fx = Fjack,x + Fjill,x + Fjane,x = ____
Fy = Fjack,y + Fjill,y + Fjane,y = ____

Direction of resultant = arctan Fy/Fx
= ____
Fill in the blanks

To find the magnitude of the net force, we need to add up the individual forces acting on the donkey.

We can break down the forces given into their x and y components based on their directions:

Jack's force (68.5 N) is acting eastward, and therefore has an x-component of 68.5 N and a y-component of 0 N.

Jill's force (61.3 N) is acting in the northeast direction, which can be split into x and y components using basic trigonometry. The angle between the northeast direction and the east direction is 45°. So, the x-component of Jill's force is 61.3 N * cos(45°), and the y-component is 61.3 N * sin(45°).

Jane's force (111 N) is acting in the southeast direction, which can also be split into x and y components using trigonometry. The angle between the southeast direction and the east direction is 45° as well. So, the x-component of Jane's force is 111 N * cos(45°), and the y-component is 111 N * sin(45°).

To find the net force, we add up the x and y components separately.

Net force in the x-direction:
= Jack's force (68.5 N, 0°) + Jill's x-component (61.3 N * cos(45°), 45°) + Jane's x-component (111 N * cos(45°), -45°)
= 68.5 N + 61.3 N * cos(45°) + 111 N * cos(45°)

Net force in the y-direction:
= Jill's y-component (61.3 N * sin(45°), 45°) + Jane's y-component (111 N * sin(45°), -45°)
= 61.3 N * sin(45°) + 111 N * sin(45°)

Now, we calculate the net force magnitude using Pythagorean's theorem:

Net force magnitude = sqrt((Net force in the x-direction)^2 + (Net force in the y-direction)^2)

To find the direction of the net force, we can use the arctan function to calculate the angle from the east direction:

Angle = arctan((Net force in the y-direction) / (Net force in the x-direction))

Substituting the values we calculated earlier will give us the magnitude of the net force and the angle from the east direction.