# Static friction on an incline

• brake4country
In summary, the coefficient of static friction between a 2 kg block and an inclined plane is approximately 0.

## Homework Statement

A 2 kg block rests on an inclined plane with an angle of 30. A force of 5 N is applied to the block in a direction down the incline plane until the block begins to move when the force reaches 7.3 N, what is the approximate coefficient of static friction between the block and the plane?

None

## The Attempt at a Solution

I made a FBD, labeled normal, weight (and its x and y components), the Ff, and the Fapplied. I made a force table to lay out all the knowns:

Fn Weight Fapplied Ff
nx=0 W=mg FAX=-7.3N Ff=μn
ny=+n Wx=-10N FAY=0N Ffx=μn
Wyy=-17.3N Ffy=0N

ΣFx=max ΣFy=may
0-10-7.3+μn = 2ax n-17.3 = 2(0)
-17.3 + μsn = 2ax n = 17.3 N

In ΣFx, I have 2 unknowns. How to solve for μs?

It says "begins to move", so you want to set a = 0 (F < 7.3 has a = 0, so F = 7.29999999999999999999999999999999999 has a = 0 -- see what I mean ?) :)

The friction co-efficient could be defined as:

The force required to break the inertia
divided by
the force due to gravity normal to the incline surface

dean barry said:
The friction co-efficient could be defined as:

The force required to break the inertia
divided by
the force due to gravity normal to the incline surface
It would be better described as stasis, not inertia.
It's not specifically to do with gravitational normal force, the source of the normal force does not matter.

OK thanks

brake4country said:
In ΣFx, I have 2 unknowns. How to solve for μs?

brake4country said:
ΣFx=max ΣFy=may
0-10-7.3+μn = 2ax n-17.3 = 2(0)
-17.3 + μsn = 2ax n = 17.3 N

Why are you considering acceleration? They have given you the limiting condition. Like BvU said, for 7.29999999999999, the acceleration is 0. So for all practical purposes, what assumption do you think you can make?