Newtons Law: Determine Friction Forces on a Box at Rest

In summary, the box will remain at rest if an applied force is not greater than the normal force, which is calculated to be 152.45 N.
  • #36
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
should it be plus mgcos 45 +p or cos45 -p?
 
  • #38
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
cause I am getting +p and it equates to 43 N
 
  • #40
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
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
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 won't have to give out your email address over the net
 
  • #43
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
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 I am actually learning and absorbing what your telling me..
 
  • #45
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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