Pre calculus HW

Question

A 75-Ib weight hangs from a string as shown in the figure. Find the tensions T1 and T2 in the string. (Round your answers to one decimal place.) T1 is 50 degrees and T2 is 30 degrees.

I can’t display a figure of the picture but the picture is T1 50 degrees to the left and T2 is 30 degrees to the right.

Answer 1

We can start by finding the vertical component of the weight, which is equal to its weight:

W = 75 lb

Next, we can find the horizontal and vertical components of T1 and T2 using trigonometry. Let's define the angles as shown in the figure:

1. T1 makes an angle of 50 degrees with the vertical, so its horizontal component is T1cos(50°) and its vertical component is T1sin(50°).
2. T2 makes an angle of 30 degrees with the horizontal, so its horizontal component is T2cos(30°) and its vertical component is T2sin(30°).

Since the weight is in equilibrium, the sum of the vertical components of T1, T2, and W must be zero:

T1sin(50°) + T2sin(30°) - W = 0

Substituting the values, we get:

T1sin(50°) + T2sin(30°) - 75 = 0

Solving for T1, we get:

T1 = (75 - T2*sin(30°)) / sin(50°)

To find T2, we can use the fact that the sum of the horizontal components of T1 and T2 must balance the weight:

T1cos(50°) - T2cos(30°) = 0

Substituting the expression for T1, we get:

(75/sin(50°) - T2*cos(30°)) * cos(50°) = 0

Solving for T2, we get:

T2 = 75/(sin(50°)*cos(30°))

Plugging in the values and rounding to one decimal place, we get:

T1 ≈ 47.4 lb T2 ≈ 57.6 lb

Therefore, the tension in the string T1 is approximately 47.4 lb and the tension in the string T2 is approximately 57.6 lb.