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Newtons Law Question

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  • #26
arildno
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CORRECT!

Now, before continuing, let us recoup what we know about the forces along the incline:

1. Frictional force Ff: We have found that this equals Ff=(mu)mgcos(45), where (mu) is the coefficient of static friction.
Further, as you have stated, Ff works downwards along the incline.

2. Gravity's component: mgsin(45), this also works downwards along the incline.

3. Applied force A: This works UPWARDS along the incline as given in the exercise.


Now, relate these quantities together in Newton's 2.law along the incline, knowing that the box is at rest!
 
  • #27
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ok.. so this is what my paper looks like

Fapp or A must not exceed ((mu)mg cos45 + mg sin 45) or the box will no longer be at rest according to newtons 2nd law. am i in the right direction?
 
  • #28
arildno
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That is correct!
And you're finished with a) now..:smile:

(OR, now you can put in the numbers if you like..)
 
  • #29
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oh thank god.. now where do i start with b.. and again i thank you soooo much you dont know what this means to me!!
 
  • #30
arildno
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Okay, about b).

Before doing any maths at all, WHY would the box start sliding if you applied a too strong perpendicular force on it?
 
  • #31
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there is not enough frictional force for it to remain stationary thus it is overtaken and the box begins to slip. In other words the weight of the box is now to great for frictional force alone to keep the box from moving
 
  • #32
arildno
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No, that is not correct:

1. You do not add any weight to the box at all by applying a perpendicular force to it. You can only add weight to the object by either moving it to another planet with stronger value for g, ior more simply, add mass to the box.

2. Furthermore, even if you did that, the normal force N, that keeps the box from going through the incline would increase correspondingly, meaning that MAXIMAL STATIC FRICTION would also increase.

The proper explanation is the following:
By pulling perpendicular on the box (not pushing!), the normal force from the incline will be reduced, since the force it now has to counteract is no longer mgcos(45), but mgcos(45)-P, where P is the force by which you pull the box.

But this means that the maximal static friction ALSO decreases, since it is proportional to the normal force!

However, in order not to slip, the friction force has to balance gravity's component along the incline, mgsin(45).

Pulling too strongly on the box lets maximal static friction below the critical value mgsin(45), and the box starts slipping.

Did you get that?
 
  • #33
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absolutely! if the book only explained things like you did i would be flying through this course.. so lets me see if i understand this..
that means that
Fsmax must be greater than mgsin 45 or it will yeild to gravitational pull along the incline? so do i know calculate at what value of p will cause fsmax to equal mgsin 45?
 
  • #34
arildno
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Correct!
The simplest way to do this in order to get your signs right, is to split your calculations in two:

First find out the magnitude of normal force necessary to keep the box from slipping (Remember, maximal friction is proportional to the magnitude of the normal force!).
Here, evidently, the critical point is when maximal static force EQUALS the gravitational pull.

Secondly, find out what this means P should be.
 
  • #35
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should i be using static friction or kinetic friction for this question?
 
  • #36
arildno
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Hmm..there is a silliness in the exercise as given I overlooked:
Since the ramp happens to be angled at 45 degrees, the components of gravity along&perpendicular to the ramp are EQUAL.
But this means that with the coefficient of friction as given (less than 1), the box CANNOT be at rest on the ramp without an additional force pushing perpendicularly down on the box. Thus, contrary to what the exercise text said, the box isn't at rest UNLESS you apply a force on it.

Thus, you'll get a minimum push on the block, rather than a minimum pull.
The thinking given above in 32, however, remains correct.


It should be static friction!
 
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  • #37
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should it be plus mgcos 45 +p or cos45 -p?
 
  • #38
arildno
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I'll do this one for you:

If positive P is pulling (i.e, away from the incline), then the box remains at rest if:
[tex]\mu(mg\cos(45)-P)=mg\sin(45)[/tex]
This is Newton's 2.law of motion along the incline when the acceleration is zero, with [itex]\mu[/itex] the coefficient of static friction.
Thus, solving for P, we get:
[tex]P=mg\cos(45)-\frac{mg\sin(45)}{\mu}[/tex]
this will be a negative number, i.e, we really have to push on the box in order to keep it in place.
 
  • #39
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cause im getting +p and it equates to 43 N
 
  • #40
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i used downward being positive which gave me +p and its just the opposite sign it still works out to be 43 N is that correct?
 
  • #41
arildno
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I haven't checked the numerical values; the sign change is because your positive p represents "push", whereas my positive P represents "pull"

(That is if you got -43 from my formula..)
 
  • #42
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yes i did.. do you have msn messenger? it would be a lot faster talking over that you can add me if you do.. so that you wont have to give out your email address over the net
 
  • #43
arildno
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Well, I'd like to keep my conversations here on PF, if you don't mind. But you can certainly use the private mail option here! :smile:

And, no, I haven't got myself MSN messenger yet. I think..
 
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  • #44
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never really though of it. are you on often, because i just completed my last lesson and i am moving on to circular motion
i really appreciate the help.. and id prefer to just communicate threw private message, i like the way you teach and im actually learning and absorbing what your telling me..
 
  • #45
arildno
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Well, that's nice to hear!
As for me being on often..click on my profile to see how many posts a day I average.
You can PM me whenever you like.

Furthermore, there are a lot of others here at PF who are good at teaching maths&physics. Welcome aboard! :smile:
 

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