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------------------------------------------------ Three people pull simultaneously on a stubborn donkey.

- Jack pulls directly ahead of the donkey with a force of [tex]98.3 \, \text{N}[/tex].
- Jill pulls with [tex]92.7 \, \text{N}[/tex] in a direction [tex]45^\circ[/tex] to the left.
- Jane pulls in a direction [tex]45^\circ[/tex] to the right with [tex]145 \, \text{N}[/tex].

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

To find the magnitude of the net force, we need to consider the vector components of the forces applied by Jack, Jill, and Jane. Jack pulls directly ahead of the donkey, so his force is entirely in the horizontal direction. Jill's force is at a 45° angle to the left, which can be split into horizontal and vertical components. Jane's force is at a 45° angle to the right, which can also be split into horizontal and vertical components.

Calculating the horizontal components:

- Jack's force: F_Jack = 98.3 N (horizontal component)

- Jill's force: F_Jill_horizontal = 92.7 N * cos(45°)

- Jane's force: F_Jane_horizontal = 145 N * cos(45°)

Calculating the vertical components:

- Jill's force: F_Jill_vertical = 92.7 N * sin(45°)

- Jane's force: F_Jane_vertical = 145 N * sin(45°)

Next, we add up the horizontal components and the vertical components separately. The net horizontal force (F_net_horizontal) is the sum of the horizontal components, and the net vertical force (F_net_vertical) is the sum of the vertical components.

F_net_horizontal = F_Jack + F_Jill_horizontal + F_Jane_horizontal

F_net_vertical = F_Jill_vertical + F_Jane_vertical

Finally, we can calculate the magnitude of the net force (F_net) using the Pythagorean theorem:

F_net = sqrt(F_net_horizontal² + F_net_vertical²)

Plugging in the values and performing the calculations, we find that the magnitude of the net force exerted by the people on the donkey is approximately 200.5 N.

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