Static friction on an incline

[brake4country]

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?

Homework Equations

None

The Attempt at a Solution

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

F_n Weight F_applied F_f
n_x=0 W=mg F_AX=-7.3N F_f=μn
n_y=+n W_x=-10N F_AY=0N F_fx=μn
W_yy=-17.3N F_fy=0N

ΣF_x=ma_x ΣF_y=ma_y
0-10-7.3+μn = 2a_x n-17.3 = 2(0)
-17.3 + μ_sn = 2a_x n = 17.3 N

In ΣF_x, I have 2 unknowns. How to solve for μ_s?

 

[BvU]

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 ?) :)

 

[dean barry]

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

 

[haruspex]

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.

 

[dean barry]

OK thanks

 

[siddharth23]

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

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?