Given a wooden sign that is held up by two ropes. One rope at a 37.0 degrees to the horizontal has a tension of 2.00N and the other at an angle of 153.5 degrees to the horizontal feels a tension of 1.785N. How much does the sign weight?

[2.00 * sin(37.0º)] + [1.785 * sin(153.5º)]

Fg=2N

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To determine the weight of the sign, we can use the concept of equilibrium. When an object is at rest, the forces acting on it must be balanced.

In this case, we have two ropes holding up the sign. The tension in each rope contributes to the vertical and horizontal forces acting on the sign.

Let's break down the forces acting on the sign:

1. Tension of Rope 1 (37.0 degrees to the horizontal): We can resolve this tension into its horizontal and vertical components. The vertical component will oppose the weight of the sign, and the horizontal component will be balanced by the horizontal component of the tension in Rope 2.

Vertical component of Tension 1 = Tension 1 * sin(37.0 degrees)

2. Tension of Rope 2 (153.5 degrees to the horizontal): Similarly, we can resolve this tension into its horizontal and vertical components. The horizontal component will balance the horizontal component of Tension 1.

Horizontal component of Tension 2 = Tension 2 * cos(153.5 degrees)

3. Weight of the sign: This is the force acting downward and can be considered the vertical component of Tension 1.

Now we can set up an equation to find the weight:

Weight of the sign = Vertical component of Tension 1 = Tension 1 * sin(37.0 degrees)

Using the given values, we have:

Weight of the sign = 2.00N * sin(37.0 degrees)

Evaluating this expression, we find that the weight of the sign is approximately 1.20N.