Calculating Coefficient of Static Friction in Simple Harmonic Motion

[hemetite]

A horizontal platform vibrates with simple harmonic motion in the horizontal directions with a perid of 2.0s. A small object place on the platform start to slide when the amplitude of vibration reachs 0.4m. Calculate the coefficient of static friction between the object and the platform.

man...i don't even know where to start...anybody can help?

what i know is that if there is no movement of the small object, we know that
fs=F


thanks a million

 

[Topher925]

Start with a FBD of a box on a surface. You know the acceleration of the surface and how this affects the box.

 

[hemetite]

what is FBD?...short form of what?

 

[Oerg]

free body diagram

 

[hemetite]

any clue to start with?...

i know i need to use the euqation of f=un

:(

 

[OwenMc]

I would start this problem by thinking about the plate. You correctly said that before the object starts to move the friction force is equal to the force exerted by the plate. You also know 2 key facts about the plates motion, its time period and its amplitude. If you have a look at the equations for SHM you will see that you can use these two things to calculate the acceleration of the plate (which is a maximum when at the extremes of motion, when the position is equal to the amplitude).
Once you know the acceleration (which you can relate to a force), you can work out the friction just like any other starts problem.

Hope this helps! I used to hate these sort of questions, but one day they just sort of click!

Cheers,
OwenMc

 

[hemetite]

this is what i get

it will start to slide when the Acting force is more than the frictional force...

mui * n = 0.4sin wt

what i know from textbook is that n= is the magnitude fo the normal force exerted by one surface on the other.

at maximum amplitude i can make sin wt= 1

hence i will get

mui * n = 0.4

now..how the find n value?

 

[Topher925]

this is what i get

it will start to slide when the Acting force is more than the frictional force...

mui * n = 0.4sin wt

what i know from textbook is that n= is the magnitude fo the normal force exerted by one surface on the other.

at maximum amplitude i can make sin wt= 1

hence i will get

mui * n = 0.4

now..how the find n value?

I think your not interpreting the equations correctly. The RHS of your first equation, 0.4*sin(wt) is the equation for the position of the wave, not the force. The 0.4 is your amplitude, w the frequency, and t, is the point in time. The way I solved this equation was to have the from the frictional force of the plate less than the force created from the acceleration of the block. First thing to do is determine the frequency of the wave. Then how do you find its maximum acceleration? What equation will give you the relation between the frictional force and the accelerative force?

Force of friction < Force from acceleration
BTW, I found a solution of mui ~= 0.402

 

[hemetite]

isnt the force is a sinusoidal osclilationg force that act on the plate...

anyway..

a=-w square A cos (wt + teta)

fs<F for it to slide off..

mui * n = m * (-w square A cos (wt + teta)

n is the normal force = fg =mg

mui * m * 9.81 = m * (-w square A cos (wt + teta)

mui * 9.81 = -w square A cos (wt + teta)

at maximum

mui * 9.81 = -(2pi/T square) A

= -0.402

am i correct?

 

[OwenMc]

Thats the same as the answer I got, except for the minus sign.
You know the harmonic force is acting in the -ve sense (trying to slow the plate down at the point of maximum acceleration), therefor the friction must be acting in the +ve sense (opposing the force from the plate). Because the object isn't moving you know the sum of the forces is zero. therefor:

Fs + Fp = 0

Or,

mui * 9.81 + (-(2pi/T square) A) = 0

So,
mui * 9.81 = (2pi/T square) A

and, mui = 0.402

 

[Topher925]

mui * 9.81 = -w square A cos (wt + teta

I think you got it, except you don't need the teta at the end since we don't care about the phase. Also, I used f = 1/T for my answer not f = 2pi/T so pay no attention to my original answer.

 

[hemetite]

thanks guys...now my confidence is growing...