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A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s= 1.5 ft.

User Colmulhall
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Answer:

attached below

Step-by-step explanation:

A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The-example-1
User Donnalee
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The concept of equilibrium which is useful when dealing with forces is states that a body on which exactly balanced forces act has no net force acting on it and the body is said to be in equilibrium

The weight of the suspended block E is approximately 18.33-lb

The reason why the value of the weight is correct is as follows:

The given parameters are;

The length of the chord = 4-ft.

The weight of the block D = 10-lb

The horizontal distance between pin A and C = 1 ft.

The length of the segment of the string s when the system is in equilibrium = 1.5 ft.

Required:

To determine the weight of the suspended block E, if the system is in equilibrium

Solution:

The tension in the cord, T = The weight of block D = 10-lb

By equilibrium of forces, we have;

E = 2 × T × cos(θ)

Where:

E = The weight of the block located at E

θ = The angle between segment CB and the vertical


sin(\theta) = (Opposite)/(Hypotenuse)

The length of the opposite side to θ = 1 ft./2 = 0.5 ft.

The length of the hypotenuse side =
(4 - 1.5)/(2) = 1.25

Therefore;


sin(\theta) = (0.5)/(1.25) = 0.4

θ ≈ 23.58°

E = 2 × 10 × cos(23.58°) ≈ 18.33 lb.

The weight of the suspended block E ≈ 18.33 lb.

Learn more about the equilibrium of forces here:

A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The-example-1
User Juan Balmori
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