How to Determine Maximum Spring Stretch for a Box on an Inclined Plane?

AI Thread Summary
To determine the maximum spring stretch for a box on an inclined plane, the maximum static friction must first be calculated using the normal force, which is influenced by the angle of inclination and the weight of the box. The static friction force can be found by multiplying the normal force by the coefficient of static friction. Once the maximum static friction force is established, it can be equated to the spring force, calculated using Hooke's Law (F = kx), where k is the spring constant and x is the stretch. The relationship allows for determining how far the spring can be stretched before the box begins to move. Understanding these forces and their interactions is crucial for solving the problem accurately.
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A 2.0kg box rests on a plank that is inclined at a angle of 65 degrees above the horizontal. The upper end of the box is attached to a spring with a force constant of 360 N/m. If the coefficient of the static friction between the box and the plank is 0.22, what is the maximum amount the spring can be stretched and the box remain at rest?

I'm lost at how to go about at the problem. I started with figuring out the max static friction but I can't figure out how that would relate to the springs force constant.
 
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If you have the max static friction you can then calculate the load required to start the box moving using the normal force, etc. This force will be in some amount of Newtons, say 200N (not correct answer). It's then a simple matter of looking at the spring and saying, "How far can I stretch the spring (or, how much load can I apply to the spring) before the box will move?"

If the required force to move was 200N and the spring rate is 360 N/m then you could stretch the spring a maximum of 200N/(360N/m) = .556m

Got it?
 
So how or what equation does the max static friction force go into the equation to figure out the normal force?
 
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