Hello there,
As you have probably guessed, i am stuck on this experiment. To cut a long story short,
my aim is to determine the coefficient of friction. So any help would be appreciated. Basically, i have a block of wood (X grams) and an inclined plane which i can adjust height. So my question is what do i do next? I am lifting the inclined plane until the block of wood begins to slide down. At the same time i record the angle that this occurs. But i don't know what to do next to find the coefficient. I knw tht F/N = μ but i can't see how that relates to my case. Please help, id be very grateful.
Thanks
Tony
A Simple Experiment, goin not so simply! Determination of the coefficient of friction
ad absurdum
Posted 08 February 2007 - 09:08 PM
Okay, well on the point of slipping:

Where
is the coefficient, N is the normal reaction of the plane on the object and F is the frictional force acting up the plane.
Now, since there is no motion, we can say that the forces are balanced. Resolving the forces perpendicular to the plane we get:
(the component of the weight acting into the plane equals the reaction force)
and resolving parallel to the plane we get:

(the frictional force up the plane is equal to the component of the weight acting down the plane)
Can you see how to use these in the equation you posted?

Where

Now, since there is no motion, we can say that the forces are balanced. Resolving the forces perpendicular to the plane we get:

(the component of the weight acting into the plane equals the reaction force)
and resolving parallel to the plane we get:

(the frictional force up the plane is equal to the component of the weight acting down the plane)
Can you see how to use these in the equation you posted?
tonyvers
Posted 08 February 2007 - 09:15 PM
Thanks for the speedy reply! yeah that helps a lot!! So just to clarify i could do μ=F/N which is the same as μ = xgsinθ/xgcosθ ?? and then jus replace with numbers. I know this is a bit simple but would i change x to the weight of the block? g to 9.8 and θ to the angle it begins to slide? Thanks again..
TOny
TOny
ad absurdum
Posted 08 February 2007 - 09:33 PM
Thanks for the speedy reply! yeah that helps a lot!!
You're welcome 
QUOTE
So just to clarify i could do μ=F/N which is the same as μ = xgsinθ/xgcosθ ?? and then jus replace with numbers. I know this is a bit simple but would i change x to the weight of the block? g to 9.8 and θ to the angle it begins to slide? Thanks again..
TOny
Yeah, you've got the right idea for x,g, and TOny


the "xg" can cancel from this fraction as it is in the numerator and the denominator, and also, the trigonometric identity


Which is quite interesting, because the angle at which it starts sliding is completely independent of the mass!
dfx
Posted 08 February 2007 - 10:09 PM

In the above formula,

