Newton's Second law: Help, please?

  • Thread starter Khamul
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Homework Statement


Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are ([tex]\mu[/tex]k)A = 0.30 and ([tex]\mu[/tex]k)B = 0.32, determine the initial acceleration of each box.


Homework Equations


[tex]\Sigma[/tex]F = m * a
m = W / g
Ffriction=[tex]\mu[/tex]*N


3. F.B.D.
6hpd90.png


4. The attempt at a solution
Hello, again, wonderful people! It's me again with another dynamic problem; although this one is centered around Newtons' second law. I am afraid to admit that I am a bit unsure where to start this problem, as the dimensions of it confuse me. Boxes A and B are touching eachother, which begs me to ask the question...are there two normal forces on one or both of these boxes?

Also to sort of go with that, how would I orient the force normals? (If there are multiple?) I realize that the angle to use is 15, but I am unsure of using sin or cos. I have always assumed that if you're orienting an angle to the horizontal(x axis) you use cos, and if you're orienting an angle to the vertical (y - axis) you use sin, but I have witnessed that this is not the case.

Could someone please help me out with my questions...as well as giving me a sort've spurn in the right direction? Thank you! :smile:

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi Khamul! :smile:

(have a mu: µ and a sigma: ∑ :wink:)
Boxes A and B are touching eachother, which begs me to ask the question...are there two normal forces on one or both of these boxes?
best to make it two

(it probably makes no difference in this case, but hey, it's only ink! :rolleyes:)
Also to sort of go with that, how would I orient the force normals? (If there are multiple?) I realize that the angle to use is 15, but I am unsure of using sin or cos. I have always assumed that if you're orienting an angle to the horizontal(x axis) you use cos, and if you're orienting an angle to the vertical (y - axis) you use sin, but I have witnessed that this is not the case.
(This is a very strange use of the word "orient" … I think you mean "resolve", ie find the component of the force, in a particular direction :wink:)

Golden rule: it is always always ALWAYS cos …

the component of a force in a particular direction is always found by multiplying by the cos of the angle between.

When you see sin (instead of cos), it's simply because they're using the "wrong angle" …

there's nothing wrong with using the "wrong angle" …

just remember that if you've labelled the angles so that the angle between this force and this direction is 90°- θ, then it's cos(90°- θ), which is sinθ. :smile:

(I always double-check by asking myself "if the plane was horizontal, would the component be zero?" … if yes, it's sin, if no, it's cos. :wink:)

(If you're worried whether it's N = gcosθ or g = Ncosθ, you'll get the correct equation by taking components ("resolving" :wink:) in the normal direction, because in that direction ∑F = ma = 0.)
 

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