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

[rasputin66]

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

 

[Doc Al]

What forces act on the board? What are the conditions for equilibrium?

 

[rasputin66]

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

 

[Doc Al]

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?

 

[asteriatic]

x= amount of stretch

To solve this problem, we can use the equation F = kx, where F is the force applied to the spring, k is the spring constant, and x is the amount of stretch. In this case, the force applied to the spring is the weight of the board, which can be calculated using the formula F = mg, where m is the mass of the board and g is the acceleration due to gravity (9.8 m/s^2). So, the force applied to the spring is:

F = (8.5 kg)(9.8 m/s^2) = 83.3 N

Now, we can plug this force into the equation F = kx and solve for x:

83.3 N = (184 N/m)(x)

x = 83.3 N / 184 N/m = 0.452 m

Therefore, the amount by which the spring is stretched is 0.452 m. This means that the spring has stretched 0.452 meters from its unstrained length in order to support the 8.5 kg board at a 50 degree angle.