Let's solve the new problem where:

• The tension in the second wire, $F_2 = 140$ N.
• The angles of the wires with respect to the horizontal: $\theta_1 = 49^\circ$ (associated with $F_1$) and $\theta_2 = 37^\circ$ (associated with $F_2$).

Step 1: Set Up the Equilibrium Equations

The equilibrium conditions for forces in both the horizontal (x) and vertical (y) directions are:

1. Horizontal (x-direction):

$F_1 \cos(49^\circ) = F_2 \cos(37^\circ)$

2. Vertical (y-direction):

$F_1 \sin(49^\circ) + F_2 \sin(37^\circ) = w$

Step 2: Solve for $F_1$

Rearrange the first equation to solve for $F_1$:

$F_1 = F_2 \frac{\cos(37^\circ)}{\cos(49^\circ)}$

Substitute the given values:

$F_1 = 140 \times \frac{\cos(37^\circ)}{\cos(49^\circ)}$

Let's compute $F_1$.The magnitude of the tension in the first wire, $F_1$, is approximately 170.4 N.

Step 3: Solve for the Weight $w$

Now, use the vertical equilibrium equation to solve for $w$:

$w = F_1 \sin(49^\circ) + F_2 \sin(37^\circ)$

Substitute the values for $F_1$, $F_2$, and the angles:

$w = 170.4 \times \sin(49^\circ) + 140 \times \sin(37^\circ)$

Let's calculate $w$.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually compute the weight $w$ using the formula:

$w = 170.4 \times \sin(49^\circ) + 140 \times \sin(37^\circ)$

Substitute the values for the sine functions using a calculator, and you will find the value for $w$.

If you have any more questions or need further help, feel free to ask!

Related Questions

1. How do different angles affect the tension in the wires?
2. What if the angles were equal; how would that impact the results?
3. How to find the tension in three wires holding a weight?
4. What are the real-life applications of such tension calculations?
5. How does changing the weight affect the force in each wire?

Tip: Understanding trigonometric functions is essential for solving problems involving forces and equilibrium.