Find Stretched Length of Spring - 8.5kg Board at 50°

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To find the stretched length of the spring supporting an 8.5 kg board at a 50° angle, the forces acting on the board, including gravity and the spring force, must be analyzed. The spring constant is 184 N/m, and the gravitational force acting on the board is calculated using the formula F = mg. The equilibrium conditions require that the sum of forces and moments equals zero, which involves considering the angle of the board. The correct approach involves resolving the forces into components and applying rotational equilibrium principles. Understanding these factors is crucial for accurately determining the spring's extension.
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A 8.5 kg board is wedged into a corner and held by a spring at a 50.0° angle, as the drawing shows. The spring has a spring constant of 184 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length.

(The picture shows a spring sticking out of the wall horizontally, and it is attached to one end of the board. The spring is holding the board up from falling flat on the floor. So the board is pulling on the spring.)

I know how to do some spring questions, but I don't know how to plug in the angle in this question. Please help! I know I need to find N to find x.


00000 <~~ spring
| / <~~~ board
| /
|/



m= 8.5 kg
angle= 50 degrees
k= 184 N/m
 
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What forces act on the board? What are the conditions for equilibrium?
 
Gravity? And the spring?

They equal zero? I think I'm missing something. Like how to figure in this angular business.

x = (mg/k)
x=(9.8x8.5)/184
x=0.565 m<~~~ not right
 
Last edited:
rasputin66 said:
Gravity? And the spring?
Yes, those are the two most important forces.
They equal zero? I think I'm missing something. Like how to figure in this angular business.
What is the condition for rotational equilibrium?
 

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