Find Coefficient of Friction for Physics Exam Problem

[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 acceleration 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 acceleration 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?

 

[arildno]

Secondly, did you regard your acceleration as a POSITIVE or NEGATIVE quantity?

 

[GuruGhulab]

1.25*9.8*sin 35

 

[GuruGhulab]

k, i think made some mistakes, let me so my calculations again, and yea acceleration used a positive value.

 

[arildno]

1.25*9.8*sin 35

Indeed, that is valid.
Now provide an answer to my second question.

 

[arildno]

k, i think made some mistakes, let me so my calculations again, and yea acceleration used a positive value.

Where is where your flaw lies.

The direction of acceleration is PARALLELL to the direction of tangential gravity, so if you assign negative direction to gravity, you must assign negative direction to acceleration.

The direction of force of friction is ANTI-PARALLELL to the direction of acceleration (trying to reduce acceleration), therefore these two quantities must have opposite signs.

 

[GuruGhulab]

k, some how i got the rite answer, but i m very confused, now, the steps i posted earlier are rite though, rite, ff, fg etc stuff.

 

[GuruGhulab]

you are rite, i made fg positive, Thanks.

 

[GuruGhulab]

Hey thanks a lot arildno, appreciate it man, actully i don even know if ur a man, anywayz whatever u r, thanks a lot, and r u really in Norway ?

 

[arildno]

A word of advice:
Structure your thoughts when doing physics; at the moment, you do not have sufficient control over your mind.

You gain that control by setting up presuppositions in an orderly manner, before attempting to solve a problem.
Also, putting in numerical values is DESTRUCTIVE of orderly, analytical thinking, because you thereby MERGE (unimportant) values from the (all-important) relations/laws into an unimportant decimal number.

It is the RELATIONS between quantities that are important when doing physics, NOT the values of those quantities as such.
Those relations, however, can only be properly stated in ALGEBRAIC form, i.e, with the aid of symbols/letters SIGNIFYING the quantities related to each other.

 

[arildno]

Hey thanks a lot arildno, appreciate it man, actully i don even know if ur a man, anywayz whatever u r, thanks a lot, and r u really in Norway ?

Yes, I am a man.
Yes, I am in Norway.