1. ## Overcoming Friction?

With regards to finding the angle (theta)

u = 0.22
m = 8

when mg*Sin(angle) = uN
and mg*Cos(angle) = N

so then I work out:

mg*sin(angle) = 0.22*mgCos(angle) ........after this i get lost,

You narrow it down to 0.22*Cos(angle) =

Tan(angle) = 0.22

I Know the answere is roughly 12.4, could someone please run through this for me, Thanks!

2. Originally Posted by bobchiba
With regards to finding the angle (theta)

u = 0.22
m = 8

when mg*Sin(angle) = uN
and mg*Cos(angle) = N

so then I work out:

mg*sin(angle) = 0.22*mgCos(angle) ........after this i get lost,

You narrow it down to 0.22*Cos(angle) =

Tan(angle) = 0.22

I Know the answere is roughly 12.4, could someone please run through this for me, Thanks!

Am I correct in saying that you have an object of mass m on a plane that is inclined at (angle) degrees and you wish to find the maximum angle such that the object does not slide? (Coefficient of static friction = u = 0.22)

If so, draw a Free-Body Diagram. I have a normal force (N) acting perpendicular to and out of the plane, a weight (w) acting directly downward, and a static friction force (f) acting up the plane. I am defining a +x direction down the plane and a +y direction directly out of the plane (in the direction of the normal force.)

Now, we are looking for the maximum possible angle, so this implies we are using the maximum static friction force. Thus
f = uN

Newton's 2nd Law in both coordinate directions:
SumFx = -f + w*sin(angle) = 0 <-- The object is stationary, thus it can't be accelerating.
SumFy = N - w*cos(angle) = 0

The top equation becomes:
-uN + mg*sin(angle) = 0

The bottom equation becomes
N = mg*cos(angle)

Inserting this value of the normal force into the top equation:
-u[mg*cos(angle)] +mg*sin(angle) = 0 <-- Cancel the common mg

-u*cos(angle) + sin(angle) = 0

u*cos(angle) = sin(angle)

sin(angle)/cos(angle) = tan(angle) = u = 0.22

Thus
(angle) = 12.407 degrees

-Dan