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?

 

[GuruGhulab]

what force would we call the force that was used to release the block, (applied force), if yes, force of friction would be equal to applied force rite, because of Newtons third law.

 

[GuruGhulab]

Sorry for my random guesses, but is the x-component, or the horizontal component, of force of gravity be equal to magnitude with force of friction.

 

[arildno]

No!
The force of friction equals u*mgcos(35), where "u" is your unkown coefficient.

 

[GuruGhulab]

yes, is my second last post any helpful.

 

[arildno]

No, it is not.

 

[arildno]

Now, set up Newton's law of motion ALONG the incline.

 

[GuruGhulab]

k, so how do we calculate force of friction?,

 

[arildno]

I've already done that; it is u*mgcos(35)

 

[GuruGhulab]

but we don't no u, and, i need some help to use Newton's second law, here, all the forces, affecting this thing are- force of gravity, force of normal, force of friction, and applied force ?

 

[arildno]

correct!
We do NOT know u yet, it is indeed what we were asked to find!
But we DO know, that whatever u is, the force of friction necessarily will be u*mgcos(35).
Agreed?

 

[GuruGhulab]

Agreed, and is the 1.84 m an helpful.

 

[arildno]

All info is helpful, in good time!

Now, once the object is moving, the only forces acting upon it in its direction of motion is the friction force and the tangential component of gravity.

What equation relates forces to acceleration?
Try to set up that equation with what you know already!

 

[GuruGhulab]

since net force is = ma
and my question is can we calculate accleration here using the equation

d=vit+ 1/2at^2.

 

[arildno]

f=ma is indeed correct.
Can you set up an expression for f now?

 

[arildno]

and my question is can we calculate accleration here using the equation

d=vit+ 1/2at^2.

That is, indeed, a good choice to determine the ACCELERATION, since you have been provided with distince, initial velocity and time.

You need that value of acceleration in order to determine "u" appearing in your f=ma equation.

 

[GuruGhulab]

is it going to be:making east positive.

f=force of friction- force of tangential..gravity.

but this is only the forces in the horizontal direction ?

 

[GuruGhulab]

i calculated acceleration, it same out to be 2.65., and BTW, the answer we should get is .37 for the coefficient.

 

[arildno]

First of all:
Those are forces in the TANGENTIAL direction, not horizontal direction.

Secondly:
You have already USED Newton's 2.law in the normal direction, namely that the normal force had to BALANCE the normal component of gravity, since the object experienced no acceleration normal to the ramp.
the information GAINED from Newton's 2law of motion in the normal direction, was that fn=mgcos(35)

 

[GuruGhulab]

Yea i mean in that tang.. direction, k , this is what i did, Ff-Fg=ma, then Ff=ma+Fg, then use the ff, in this equation, u=Ff/Fn. but got the wonrg answer.

 

[arildno]

What did you use for Fg?