A 171-N child sits on a light swing and is pulled back and held with a horizontal force of 73.5N. The magnitude of the tension force of each of the two supporting ropes is?

Let T be the single-rope tension and A be the angle from vertical

2T sinA = 73.5
2T cosA = Weight = 171

Square and add both equations

4T^2 = 73.5^2 + 171^2

Solve for T

186.13

546545

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the answer is 94. I don't know how to do the problem though.

To find the magnitude of the tension force in each of the two supporting ropes, we can use Newton's laws of motion. The forces acting on the child are the weight (mg) pointing downwards and the horizontal force (F) pulling the child back. The tension forces of the supporting ropes provide the vertical component of the net force.

First, let's determine the weight of the child. We are given that the child has a weight of 171 N. The weight of an object can be calculated using the equation:

Weight (W) = mass (m) x acceleration due to gravity (g)

Given weight (W) = 171 N

Using the equation, we can solve for mass (m):

171 N = m x 9.8 m/s^2 (acceleration due to gravity)

m = 171 N ÷ 9.8 m/s^2

m ≈ 17.45 kg

Now that we know the mass of the child is approximately 17.45 kg, we can determine the vertical component of the net force. The vertical component of the net force is equal to the tension force in each supporting rope:

Vertical component = Weight of the child (mg) - Force pulling back (F)

Vertical component = 17.45 kg x 9.8 m/s^2 - 73.5 N

Vertical component ≈ 171 N - 73.5 N

Vertical component ≈ 97.5 N

Since the swing is in equilibrium (not accelerating vertically), the tension forces in the supporting ropes are equal and opposite. Therefore, each supporting rope exerts a tension force of approximately 97.5 N.