# thermodynemics?

The unstretched length of the spring is r. When pin P is in an arbitrary position theta, determine the x- and y-components of the force which the spring exerts on the pin. Evaluate your general expressions for r =400 mm, k = 1.4 kN/m, and theta = 40 degrees. (Note: The force in a spring is given by F= δk, where δ is the extension from the unstretched length.)

Relevance
• force in the spring is

F = kδ

where δ is the elongation of the spring

for an angle Θ the vertical (down) and horizontal (x) displacement of the pin is

x = r*sinΘ

y = r*(1-cosΘ)

Then the length of the spring has y-component

Y = r + r*(1 - cosΘ) = r*(2 - cosΘ)

and x-component

x = r*sinΘ

(from above)

making the length of the spring

L = √(Y² + x²) = √[r²*(2-cosΘ)² + r²sin²Θ]

L = r*√(4 - 4cosΘ + cos²Θ + sin²Θ) = r*√(5 - 4cosΘ)

which in turn makes

δ = L - r = r*[√(5 - 4cosΘ) - 1]

(As a check, we note that for Θ = 0º we have

δ = r*[√(5-4) - 1] = r*0 = 0

)

So the magnitude of the force is

|F| = kδ = k*r*[√(5 - 4cosΘ) - 1]

The angle the spring makes above the -x axis is

φ = arctan(Y/x) = arctan(r(2 - cosΘ)/(r*sinΘ)

φ = arctan((2-cosΘ)/sinΘ)

so the x-component of the force is

Fx = k*r*[√(5 - 4cosΘ) - 1]*cosφ ← to the left

and the y-component is

Fy = k*r*[√(5 - 4cosΘ) - 1]*sinφ ← up

for φ = arctan((2-cosΘ)/sinΘ)

It doesn't look to me like this can be simplified. • Login to reply the answers
• Anonymous
2 months ago

1) Not thermodynamics.

2) Diagram missing.

Well done!

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