Coefficient of maximum static friction

[xmflea]

this is a lab that I've been working on for 3 days alredy, and still cannot fill in the blanks for finding the coefficient of maximum static friction...

a wooden block was attatched to a pulley connected to a hanger.. the wooden block was on top of a wooden board.. the goal was to add just enough weight to the hanger to make the wooden block slide... the wooden block was sliding horizontally.

for one of my results... the mass of my wooden block was 0.543kg..the mass on the hanger required to move the block was 0.170kg... now i calculated the weight of the block to be 5.32N, and the weight of the hanger to be 1.67N...
now i am asked to find the coefficient of the maximum static friction... and I've tried everything for three days. multiplying and dividing up and down etc... the only equation given is Fs is less than or equal to UsN (U=mu you know that greek symbol). and they want me to find Usmax with that equation. my Lab TA has extremely broken english and when he explains he just expects me to know what he means. the lab instructor is just at the school for research and is never around to care about helping students. a friend of mine told me to find the normal force, which i tried doing by adding the two masses and multiplying by 9.8. but i don't know where to go from there.

 

[Doc Al]

Let's start by getting some terminology straightened out. μs is the coefficient of static friction. The maximum value of static friction is given by Fs = μsN. There's no coefficient of "maximum" static friction.

the only equation given is Fs is less than or equal to UsN (U=mu you know that greek symbol).

That's the only equation you need.

Here's the idea. You keep piling weight on the hanger, which pulls the block via tension in the string, but the opposing static friction is enough to prevent any motion. But when you put just enough weight on the hanger, and the tension just barely exceeds the maximum value of static friction, the block just starts to move. That's when you can apply that equation. What's the normal force on the block? What must the friction force equal?

 

[xmflea]

so the normal force on the block is 5.73N. so would the equation look like.

5.73N= μs1.67 ? and then i solve for us?

 

[Doc Al]

so the normal force on the block is 5.73N. so would the equation look like.

5.73N= μs1.67 ? and then i solve for us?

Yes, except how did you get 5.73 N?

Edit: You have the normal force and the friction force mixed up.

 

[xmflea]

5.73N is the weight of the block on the board it was found by multiplying 0.543 by 9.8.

 

[Doc Al]

5.73N is the weight of the block on the board it was found by multiplying 0.543 by 9.8.

Redo that calculation.

 

[xmflea]

so the coeficcient of maximum static friction is 3.186? μsmax = 3.186?
that number seems large

 

[Doc Al]

so the coeficcient of maximum static friction is 3.186? μsmax = 3.186?
that number seems large

Yeah, it's way off. You had the equation backwards (I mustn't have been paying attention earlier ):

so the normal force on the block is 5.73N. so would the equation look like.

5.73N= μs1.67 ? and then i solve for us?

You have the normal force and the friction force mixed up.

 

[xmflea]

so its the other way around.. i got usmax=0.314

 

[Doc Al]

so its the other way around.. i got usmax=0.314

Good. But don't call it usmax, it's just us.

 

[xmflea]

well the lab calls it usmax, but i got you , thanks a lot.

 

[Doc Al]

well the lab calls it usmax

Then you'd better use their terminology, even though it makes no sense!

And you're welcome.