Physics cord force

Question

https://drive.google.com/file/d/1W5uJY9v1bJS4EsR1yFjxK1WDwZ84k-hg/view?usp=sharing

Pull Force = 40 lb, coefficient stiffness is 35 lg/ft. Length Cord AB is d =1.5 ft long. The angle is θ = 33 degree.

1. What is the stretched length of the spring in feet

2. What is angle α in degrees

3. What is force of the spring in pounds

4 .What is unstreched length of the spring in feet

Answer 1

Let

P = 40 lb be the force pulling the string down,

F (to be calculated) be the force of the spring,

T (unknown) be the tension in the section of the string between points A and B,

d = 1.5 ft be the length of AB as indicated,

x (to be calculated) be the unstreched length of the spring,

θ = 33° be the angle indicated,

k = 35 lb/ft be the stiffness of the spring,

D be the point of intersection between the line BC and the vertical line through point A.

1. We are to calculate the length AC.

We have two right angle triangles -- ABD and ACD.

From the triangle ABD we can write

AD = dsin(θ)

BD = dcos(θ)

Therefore CD = BC - BD = 2d - dcos(θ) = d(2 - cos(θ))

From the triangle ACD using the Pythagorean Theorem

AC = √(AD² + CD²)

= √((dsin(θ))² + (d(2 - cos(θ)))²)

= d√(sin²(θ) + (2 - cos(θ))²)

= d√(sin²(θ) + 4 - 4cos(θ) + cos²(θ))

= d√(5 - 4cos(θ))

Substituting actual numbers

AC = 1.5√(5 - 4cos(33°)) = 1.924 ft

2. We are to calculate the angle α.

It is also the angle ACD.

Therefore

sin(α) = AD/AC

= dsin(θ)/d√(5 - 4cos(θ))

= sin(θ)/√(5 - 4cos(θ))

Substituting actual numbers

α = arcsin(sin(33°)/√(5 - 4cos(33°))) = 25.1°

3. We are to calculate the force F.

Assuming that the picture describes an equilibrium situation,

the vector sum

P + F + T = 0.

Therefore the projections of the three forces on the horizontal and vertical axis all add up to 0.

In an abuse of notation, let P, F, T be the magnitudes of the forces, as opposed to the force vectors.

Then the three forces must be in balance along the two axes.

Horizontal axis projection: Fcos(α) = Tcos(θ)

Vertical axis projection: P = Fsin(α) + Tsin(θ)

From the horizontal projection: T = Fcos(α)/cos(θ)

Plug that into the vertical projection condition:

P = Fsin(α) + Fcos(α)sin(θ)/cos(θ)

F = P/(sin(α) + cos(α)tan(θ))

Substituting actual numbers

F = 40/(sin(25.1°) + cos(25.1°)tan(33°)) = 39.5 lb

4. We are to calculate x.

By the definition of stiffness k

F = k(AC - x)

x = AC - F/k

Substituting actual numbers

x = 1.924 - 39.5/35 = 0.79 ft