Box on a slanted plank with friction attached to a spring

[rakailee]

Homework Statement

A 2.0 kg box rests on a plank that is inclined at an 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 static friction between the box and the plank is 0.22, what is the maximum amount the spring can be stretched and the box to remain at rest?

Relevant Equations

Fnet = ma, Ff = Fn(.22), F = kx

I first find the force of friction to be (2)(9.8)cos(65)(.22), then I find the pull of gravity to be (2)(9.8)sin(65).

The full equation I set up to be: 0 = kx + force of friction minus the pull of gravity

This gives me the wrong answer, 0.44 . My free-body diagram is that kx and force of friction go in the same direction upwards and the pull of gravity counters that. I think if I played around with the signs I would arrive at the right answer, but I can't see the logic behind it. Could someone explain?

 

[etotheipi]

My free-body diagram is that kx and force of friction go in the same direction upwards

I think your confusion is arising because, in the absence of the tension force, the friction would indeed point up the slope for equilibrium. Can you reason as to what the direction of the friction would be in this problem?

The best way to think about it is to consider what would happen if the frictional force weren't there at all - which direction would the slippage be in? Then, insert the frictional force in the direction opposite to this.

 

[rakailee]

Is “point up the slope” the same direction as the reverse spring force? I believe the slippage to be down the slope, so the fictional force would be up the slope, which is the same direction as the spring force. Is this not correct?
Edit: homework prompt should b spring, not string

 

[etotheipi]

We're interested in the point at which the tension is a maximum. Which direction does the frictional force need to point in if we still want the equilibrium condition, with the largest possible tension?

To put it another way, the box is on the point of moving up the slope...

 

[PeroK]

Is “point up the slope” the same direction as the reverse spring force? I believe the slippage to be down the slope, so the fictional force would be up the slope, which is the same direction as the spring force. Is this not correct?

To put it more blunty: if you try to pull a load up a slope, then in your model friction helps! If you get a large enough friction force, then you wouldn't have to pull at all! Friction would do all the work against gravity.

The moral is that friction always acts against motion. If the gravity force is greater than the spring force, then friction acts against downward acceleration. And, if the spring force is greater, then friction acts against upwards acceleration.

 

[haruspex]

friction always acts against motion

In other circumstances that can lead to a wrong application. Better to say it acts against relative motion of the surfaces in contact.