Calculating Net Force on a Stubborn Donkey

What is the magnitude of the net force exerted on the stubborn donkey by Jack, Jill, and Jane? What is the direction of the net force? The magnitude of the net force exerted on the stubborn donkey is approximately 247.60 N. The direction of the net force is approximately -18.56° to the right.

When three people pull simultaneously on a stubborn donkey, the net force they exert must be calculated to determine the overall effect on the donkey. In this scenario, Jack pulls directly ahead of the donkey with a force of 63.1 N, Jill pulls with 64.5 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 167 N.

To find the magnitude of the net force, we need to calculate the components of each force in the x and y directions. The x-component is the force multiplied by the cosine of the angle, and the y-component is the force multiplied by the sine of the angle. Adding up the x-components and y-components separately, we can then use the Pythagorean theorem to find the magnitude of the net force.

The x-components are: 63.1 N * cos(0) + 64.5 N * cos(45°) + 167 N * cos(-45°) = 63.1 N + 45.54 N + 117.98 N = 226.62 N.

The y-components are: 0 + 64.5 N * sin(45°) - 167 N * sin(-45°) = 45.54 N - 117.98 N = -72.44 N.

Using the Pythagorean theorem, the magnitude of the net force is √(226.62 N^2 + (-72.44 N)^2) = √(61315.32 N^2) ≈ 247.60 N.

To find the angle of the net force, we can use the inverse tangent function: θ = tan^(-1)(-72.44 N / 226.62 N) ≈ -18.56°.

Therefore, the magnitude of the net force exerted on the stubborn donkey by Jack, Jill, and Jane is approximately 247.60 N, and the direction of the net force is approximately -18.56° to the right. Despite the uncoordinated efforts of the people involved, the donkey is subjected to a significant net force in a specific direction.

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