What is the maximum spring stretch for a resting box on an inclined plank?

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To determine the maximum spring stretch for a box resting on an inclined plank, the forces acting on the box must be analyzed, including gravitational force, normal force, frictional force, and spring force. The gravitational force acting down the incline is calculated using the box's weight and the sine of the incline angle, while the normal force is determined using the cosine of the angle. The static friction force is then found using the coefficient of static friction and the normal force. The spring force is represented by F = -kx, where k is the spring constant and x is the stretch. The maximum stretch occurs when the spring force equals the sum of the gravitational force down the incline and the frictional force opposing it, leading to the equation kx = mg sin(θ) + μN.
Lnav1982
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I am having problems with this question:

An m = 2.12 kg box rests on a plank that is inclined at an angle of q = 64.5° above the horizontal. The upper end of the box is attached to a spring with a force constant of 15.2 N/m.

If the coefficient of static friction between the box and the plank is 0.243, what is the maximum amount the spring can be stretched and the box remain at rest?

Thank you
 
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You need to show some work or describe what you are thinking or where you are stuck.:wink:
 
Ok, well I tried using F=-Kx and solve for x but that didnt work becuase I didnt use the other values. Then using Fs=usN after finding N which I think was = mgcos(theta) to find Fs, then plug that into F=-kx. But this didnt work as well. What am I doing wrong? Thanks
 
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