What is the Minimum Coefficient of Friction to Prevent Block Movement?

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Minimum coefficient of friction b/t block and surface that would be required to keep block from moving?

I first solved to find the acceleration w/o friction since friction wasn't given in the problem. How do I solve for the min friction to keep the block from moving? Do I set the block's acceleration to zero and factor in static friction and then solve?

Hopefully you guys know what I mean..if not let me know
 
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I'm assuming you are given a force that will be acting on the block, and the blocks mass?
Remember that [tex]F_{f}=\mu_{s}N[/tex]. You hopefully know both the normal force and the frictional force required to counter whatever force is being exerted on the block. Can you solve for the coefficient of friction?

Is that what you were looking for?
 


There's a tension on the block, normal force, and gravity...but the problem doesn't mention a friction. So I first solved the problem for acceleration without friction at all. But then the problem proceeds to ask what coefficient of friction I would need to prevent the block from moving at all.

At first I did:

T = max

since tension is the only thing acting on the block

I didn't include friction since problem doesn't give friction.

Now to get the coefficient of friction to keep it from moving would I do this?

-Fs + T = 0?

Fs = usN being the static friction

and solve from there?

thats all i need to know
 
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Couldn't I also set The Tension equal to the weight times the friction and solve for friction? I solved it both ways(this way and the way I explained in my previous post) and got the same answer of 1.02 as my coefficient of static friction.

The tension = 100N

the w = mg

=(10kg)(9.81m/s^2) = 98.1N

100N = (98.1N)(Fs)

= 1.02Doesn't this seem a little to high though? The weight of the block is 10kg and the weight of the mass hanging by a pulley on it is 100N = 10.2 kg. You would think not much friction is required to keep it from moving...I'm not sure...any input?
 


You essentially did the same thing in both methods no?

The required coefficient is proportional to the tension force.