A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. The hoist weighs 340 N. The ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. Find the tension in each rope and the magnitude of each tension. (LetT2 and T3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively.
the horizontal components of the tensions are equal and pull in opposite directions
the vertical components of the tensions sum to the weight of the hoist
the longer rope makes the smaller angle with the horizontal
T2 * cos(50º) = T3 * cos(38º)
[T2 * sin(50º)] + [T3 * sin(38º)] = 340 N
solve the system of equations for T2 and T3
To find the tension in each rope, we can use the concept of forces in equilibrium. Since the hoist is suspended and not moving, the sum of the forces in the vertical direction and the sum of the forces in the horizontal direction must be zero.
First, let's calculate the vertical component of the tension in each rope. The weight of the hoist is acting straight down and can be defined as:
F_vertical = weight of the hoist = 340 N
Next, let's find the horizontal component of the tension in each rope. We can use the angle and trigonometry to determine the horizontal component of the tension in each rope.
For the rope of length 2 m and angle 50°, the horizontal component can be calculated as:
T2_horizontal = T2 * cos(50°)
Similarly, for the rope of length 3 m and angle 38°, the horizontal component can be calculated as:
T3_horizontal = T3 * cos(38°)
Now we can set up equations for the vertical and horizontal forces in equilibrium:
In the vertical direction: T2_vertical + T3_vertical - F_vertical = 0
In the horizontal direction: T2_horizontal + T3_horizontal = 0
Substituting the equations with appropriate values, we get:
T2_vertical + T3_vertical - 340 N = 0
T2 * cos(50°) + T3 * cos(38°) = 0
To solve these equations simultaneously, we need one more equation. We can use the fact that the rope lengths are 2 m and 3 m to write the equation:
T2_vertical / T3_vertical = 2 m / 3 m
Now, let's solve the equations:
T2_vertical + T3_vertical = 340 N (Equation 1)
T2 * cos(50°) + T3 * cos(38°) = 0 (Equation 2)
T2_vertical / T3_vertical = 2 / 3 (Equation 3)
From Equation 3, we can write T2_vertical = (2/3) * T3_vertical
Substituting this value in Equation 1, we get:
(2/3) * T3_vertical + T3_vertical = 340 N
(5/3) * T3_vertical = 340 N
T3_vertical = (340 N) * (3/5)
T3_vertical = 204 N
From Equation 3, we can now find T2_vertical:
T2_vertical = (2/3) * T3_vertical
T2_vertical = (2/3) * 204 N
T2_vertical = 136 N
To find the magnitude of each tension, we can use the Pythagorean theorem:
T2 = sqrt((T2_horizontal)^2 + (T2_vertical)^2)
T3 = sqrt((T3_horizontal)^2 + (T3_vertical)^2)
Calculating the magnitudes:
T2 = sqrt((T2_horizontal)^2 + (T2_vertical)^2)
T2 = sqrt((136 N)^2 + (T2_horizontal)^2)
T3 = sqrt((T3_horizontal)^2 + (T3_vertical)^2)
T3 = sqrt((T3_horizontal)^2 + (204 N)^2)
To find the tensions in each rope and their magnitudes, we can use the principles of equilibrium and decompose the weight of the hoist into its vertical and horizontal components.
Let's start by finding the horizontal and vertical components of the weight.
The vertical component can be found using the sine function:
Vertical Component = Weight * sin(angle)
Vertical Component = 340 N * sin(38°)
Vertical Component = 340 N * 0.61566
Vertical Component ≈ 209.42 N
The horizontal component can be found using the cosine function:
Horizontal Component = Weight * cos(angle)
Horizontal Component = 340 N * cos(38°)
Horizontal Component = 340 N * 0.78801
Horizontal Component ≈ 267.73 N
Now that we have the horizontal and vertical components of the weight, we can find the tensions in each rope.
For the rope of length 2 m:
The tension vector (T2) can be decomposed into its vertical and horizontal components. Let T2v represent the vertical component and T2h represent the horizontal component.
Using the triangle formed by the rope, vertical component, and horizontal component, we can write:
T2v = T2 * sin(50°)
T2h = T2 * cos(50°)
Since the vertical component must balance the weight's vertical component, we can equate them:
T2v = Vertical Component (209.42 N)
T2 * sin(50°) = 209.42 N
Solving for T2, we get:
T2 = 209.42 N / sin(50°)
T2 ≈ 271.36 N
Similarly, with the horizontal component, we can equate them:
T2h = Horizontal Component (267.73 N)
T2 * cos(50°) = 267.73 N
Solving for T2, we get:
T2 = 267.73 N / cos(50°)
T2 ≈ 182.36 N
For the rope of length 3 m:
The tension vector (T3) can be decomposed into its vertical and horizontal components. Let T3v represent the vertical component and T3h represent the horizontal component.
Using the triangle formed by the rope, vertical component, and horizontal component, we can write:
T3v = T3 * sin(38°)
T3h = T3 * cos(38°)
Since the vertical component must balance the weight's vertical component, we can equate them:
T3v = Vertical Component (209.42 N)
T3 * sin(38°) = 209.42 N
Solving for T3, we get:
T3 = 209.42 N / sin(38°)
T3 ≈ 346.12 N
Similarly, with the horizontal component, we can equate them:
T3h = Horizontal Component (267.73 N)
T3 * cos(38°) = 267.73 N
Solving for T3, we get:
T3 = 267.73 N / cos(38°)
T3 ≈ 337.12 N
Therefore, the tension in each rope is approximately T2 ≈ 271.36 N for the rope of length 2 m, and T3 ≈ 346.12 N for the rope of length 3 m.