Find Coefficient of Friction for Physics Exam Problem

[GuruGhulab]

Hi there, My physics exam is on monday, and i have reviwed a lot, but i can't get these last five questions, i ll post em one by one, so first i need help with the first one,

Homework Statement

A 1.25 kg block is released from rest on a ramp tilted 35 degrees above the horizontal, if the block moves 1.84m down the ramp in 1.18s, find the coefficient of friction between the block, and the ramp.

Plz help me out.
Thanks

 

[radou]

Any thoughts on the question?

 

[arildno]

What equations do you think might come in handy here?

 

[GuruGhulab]

uhh, i m think Ff=uFn, and i m not sure, if i should solve it, like a projectile, question.Yep, i m very confused.

 

[arildno]

That is, indeed an important equation!

Now, however, there is an even more important equation that is ALWAYS used in classical dynamics/kinematics problems.
Which one is that?

 

[GuruGhulab]

net force= ma, at least i think so man. Thanks.

 

[arildno]

Correct!

Remember that the direction of the force of gravity is neither parallell to the direction of motion, nor perpendicular to it.

 

[GuruGhulab]

Yes, is it coming straight down the ramp, and the angle its one of the component make is 35 degrees, i hope i m clear, Thanks again.

 

[arildno]

Note:
You should first figure out what the normal force is, expressed by the object's mass, the angle, and g.

 

[GuruGhulab]

normal force, is going to be the hypotenuse, of its components rite,and is force of gravity equal to the y-component of the normal force, if yes, then normal force would = mg/cos angle.

 

[arildno]

Not at all.

The normal force cancels the normal component of gravity, since the object experience no acceleration INTO the ramp, only ALONG it.

Therefore, the force of GRAVITY itself, mg, should be regarded as the hypotenuse in a right-angled triangle with normal&tangential components as the cathetes.

 

[GuruGhulab]

k, so the normal force cancels the Y-component of fg, rite , and is , Net force in the y-direction, zero or not. Thanks.

 

[arildno]

What do you mean by Y-component of fg?

 

[arildno]

Decompose mg in the tangential and normal directions.

 

[GuruGhulab]

k, check out the diagram and let me know, if u don;t get something. Thanks

 

[arildno]

I get it perfectly, but I don't think you do.
At the very least, what you've posted so far is wrong.

Now, look at it this way, then:
If the TANGENTIAL direction makes a 35 degree angle with the horizontal, what is the angle between the NORMAL direction and the vertical?

 

[GuruGhulab]

k let me c, one sec., and TANGENTIAL direction is the one it is heading ?, the angle makes 55 degrees, i guess.

 

[arildno]

Quite so; tangential means ALONG the ramp, normal direction means the direction head-on the ramp, making a 90 degrees angle with the tangential direction.

 

[GuruGhulab]

so what would force of normal equal.

 

[arildno]

EDIT: 55 degrees is wrong!
Please make new posts, rather than fill old posts with new stuff you are uncertain of.

 

[arildno]

Again:
What is the angle between the normal&vertical, when the angle between the tangential&horizontal directions is 35 degrees.

Please, can't you make a drawing on a piece of paper to find that out?

 

[GuruGhulab]

i m sorry, the way i m seeing it , 125, uhh which vertical r u talkin about, thanks.

 

[GuruGhulab]

if u mean the vertical of normal, then it would be 35 because of the z pattern.

 

[arildno]

Can't you see that the corresponiding angle between the normal direction and the vertical direction is also 35 degrees?

EDIT:
Seems you've figured that out by yourself now.

 

[GuruGhulab]

lolz, then u are talkin about the same angle i mean, Thanks.

 

[arildno]

So!
Now that you know the angle between the normal and vertical is 35 degrees, what is then the normal force equal to?

 

[GuruGhulab]

Please don't kill me for this, fg*sin 35, or fg cos 35, is it fg*cos 35.

 

[arildno]

It is fgcos(35).

The easy way to remember wheter you should use sine or cosine, is to see what the result should be if the angle was 0.
In the case of angle=0, the normal force is of course equal to mg=mg*1=mg*cos(0)

 

[GuruGhulab]

k, we got force of normal, but we still, need force of friction. to find the coefficient.

 

[arildno]

Yes, from your first equation, what must the force of friction equal?