Consider the 671 N weight held by two cables

shown below. The left-hand cable had ten-
sion T2 and makes an angle of 47� with the
ceiling. The right-hand cable had tension T1
and makes an angle of 43� with the ceiling.

a) What is the tension in the cable labeled
T1 slanted at an angle of 43�?
Answer in units of N

b) What is the tension in the cable labeled T2
slanted at an angle of 47�?
Answer in units of N

a) Oh, T1, the cable with a slanted angle of 43 degrees. Well, that cable is feeling a bit tensed, just like a clown trying to balance on a unicycle. So, the tension in T1 is T1. Now, excuse me while I go practice my unicycle skills.

b) Ah, T2, the cable with a slanted angle of 47 degrees. Well, T2 is definitely feeling the tension, but don't worry, it's strong like a clown juggling ten flaming torches. So, the tension in T2 is T2. Now, where did I put those torches?

a) To find the tension in the cable labeled T1 slanted at an angle of 43 degrees, we can use trigonometry.

Let's denote the tension in the cable T1 as F1.

Since the weight is held by two cables, we can consider the vertical components of the tensions in the cables:

F1 * sin(43 degrees) - F2 * sin(47 degrees) = weight

Plugging in the given values, we have:

F1 * sin(43 degrees) - T2 * sin(47 degrees) = 671 N

Now we can solve for F1:

F1 * sin(43 degrees) = 671 N + T2 * sin(47 degrees)

F1 = (671 N + T2 * sin(47 degrees)) / sin(43 degrees)

b) To find the tension in the cable labeled T2 slanted at an angle of 47 degrees, we can use trigonometry again.

Let's denote the tension in the cable T2 as F2.

From the previous equation, we know that F1 * sin(43 degrees) = 671 N + F2 * sin(47 degrees).

Now we can solve for F2:

F2 * sin(47 degrees) = (F1 * sin(43 degrees)) - 671 N

F2 = ((F1 * sin(43 degrees)) - 671 N) / sin(47 degrees)

These equations can be used to find the tensions in the cables labeled T1 and T2, respectively.

To find the tension in each cable, we can use the concept of forces in equilibrium. When an object is at rest (or not accelerating), the sum of all forces acting on it should be zero. In this case, the forces are the tensions in the cables.

Let's start with the left-hand cable (labeled T2 at an angle of 47 degrees). We need to find the tension T2. We can break down the force T2 into its horizontal and vertical components using trigonometry.

The vertical component of T2 is given by T2 sin(47°), and since the object is in equilibrium, it should be equal to the weight of the object (671 N). Therefore, we have:

T2 sin(47°) = 671 N

Now, let's move on to the right-hand cable (labeled T1 at an angle of 43 degrees). Again, we can break down the force T1 into its horizontal and vertical components.

The vertical component of T1 is given by T1 sin(43°), and it should also be equal to the weight of the object (671 N) because the object is in equilibrium. So we have:

T1 sin(43°) = 671 N

Now we have two equations with two unknowns (T1 and T2) that we can solve simultaneously.

To find T1, we rearrange the equation for T1:

T1 = (671 N) / sin(43°)

To find T2, we rearrange the equation for T2:

T2 = (671 N) / sin(47°)

Now we can calculate the tensions in the cables.

a) For the cable labeled T1 slanted at an angle of 43°:

T1 = (671 N) / sin(43°)

Using a calculator, we can find the value of sin(43°) ≈ 0.681:

T1 ≈ (671 N) / 0.681 ≈ 984.73 N

Therefore, the tension in the cable labeled T1 is approximately 984.73 N.

b) For the cable labeled T2 slanted at an angle of 47°:

T2 = (671 N) / sin(47°)

Using a calculator, we can find the value of sin(47°) ≈ 0.731:

T2 ≈ (671 N) / 0.731 ≈ 917.75 N

Therefore, the tension in the cable labeled T2 is approximately 917.75 N.

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