Two artifacts in a museum display are hung from vertical walls by very light wires. Wire T1 is horizontal from the left wall and goes to a center weight of 177N and wire T2 is at a 53 degree angle with the right wall heading from right to left downwards towards the center weight of 177N. From the center weight a wire T3 hangs vertically to a 40kg object.

Find the tension on T1, T2 and T3.
T3 solved but don't know how to calc. T1 and T2.

To find the tension in wires T1 and T2, we can use the principles of equilibrium. In an equilibrium situation, the sum of the forces acting on an object is zero.

Let's start by resolving the forces acting on the center weight:

1. Resolve the weight of the center object into horizontal and vertical components:
- Vertical component (downwards) = Weight * sin(90°) = 177 N * sin(90°) = 177 N
- Horizontal component (leftwards) = Weight * cos(90°) = 177 N * cos(90°) = 0 N

2. Resolve the weight of the 40 kg object into horizontal and vertical components:
- Vertical component (downwards) = Weight * sin(90°) = 40 kg * 9.8 m/s^2 = 392 N
- Horizontal component = 0 N (since the wire is vertical)

3. To ensure equilibrium, the sum of the vertical forces acting on the center weight must be zero:
T1 + T2 - 177 N - 392 N = 0

Now, let's consider the horizontal forces acting on the center weight:

4. The horizontal forces acting on the center weight must also sum to zero:
T1 - T2 + 0 N = 0

We now have a system of two equations and two unknowns:

T1 + T2 = 177 N + 392 N ...(Equation 1)
T1 - T2 = 0 N ...(Equation 2)

To solve for T1 and T2, add Equation 1 and Equation 2:

2T1 = 177 N + 392 N
2T1 = 569 N

T1 = 569 N / 2
T1 = 284.5 N

Substitute the value of T1 into Equation 1 to find T2:

284.5 N + T2 = 569 N
T2 = 569 N - 284.5 N
T2 = 284.5 N

Therefore, the tension in wire T1 is approximately 284.5 N, and the tension in wire T2 is also approximately 284.5 N.

To calculate the tensions T1 and T2, we can start by analyzing the forces acting on the center weight of 177N. We can break this force into its vertical and horizontal components.

1. Horizontal components:
Since the weight is hung from a horizontal wire T1, the tension T1 supports the entire horizontal weight. Therefore, the tension T1 is equal to the weight of the center wire, which is also 177N.

T1 = 177N

2. Vertical components:
The force due to the vertical weight of 177N is distributed between two wires: T2 and T3.

First, let's find the vertical component of tension T2.
The weight is connected to wire T2 at a 53-degree angle. We need to find the vertical component of T2, which can be calculated using trigonometry.

The vertical component of T2 = T2 * sin(53°)

Next, let's find the vertical component of tension T3.
Since wire T3 hangs vertically, the entire weight is supported by the tension T3. The vertical component of T3 is equal to the weight of the 40kg object, which can be calculated using the formula:

Weight = mass * gravity
Weight = 40kg * 9.8 m/s²

Now, let's sum up the vertical components:

Vertical component of T2 + Vertical component of T3 = Weight

(T2 * sin(53°)) + (40kg * 9.8 m/s²) = 177N

Now, you can rearrange this equation to solve for the tension T2:

T2 = (177N - (40kg * 9.8 m/s²)) / sin(53°)

Once you have calculated the tension T2 using this equation, you can substitute it back into the equation for T3's vertical component to calculate T3.

I hope this helps you calculate the tensions T1, T2, and T3 in the museum display!

ok, on T2, T3...

Sum horizontal forces=zero
T2-T3cos37=0
that is one equation. Second equation is sum of vertical forces is zero..
T3Sin37-177=0