anonymous
  • anonymous
An object with mass m1=4.00kg and a box with mass m2=5.00kg are attached by a wire (massless) that pass trough a sheave (negligible mass too). The object m1 is pendant and the box is in an inclined plan (sorry, there's a image but i don't know how to post it here). They are abandoned from rest. the friction coefficient between the box and the plan is both 0.2 for static and dinamic modules. sen@=0.574, cos@=0.820 and g=10m/s² a.The value of the acceleration a b.The tensor force on the wire Sorry for my bad english.
Physics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I'm having trouble to explain why the results of question a are conflictants if I consider the box going up and the box going down. The answer given was a=0.344m/s², wich I only find when the box goes up.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
What does the word "sheave" mean? Is it a pulley? Use the "Draw" button to give us a small sketch.
anonymous
  • anonymous
yes, pulley haha

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anonymous
  • anonymous
|dw:1353263124656:dw| here
anonymous
  • anonymous
That's actually a common question. I just can't tell for sure why the results are different dependending on which case i choose
anonymous
  • anonymous
and "please". I forgot to write it :P
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Imagine there is no friction: the system will move to one direction only. There is no chance that, adding friction, it may move the other way. So there is only one direction to consider when you solve the problem.
anonymous
  • anonymous
That's the point. But since I don't know the direction of the movement, wouldn't it be valid for me to randomly pic any direction? So if the acceleration results in a negative value, it just means I have choosen the wrong way. If I solve the question by the hipotesis that the box is going up, the Answer i get is indeed a=0.344444...m/s². But if I try for the oposite, That means the frictional force must point to the other direction, so it doesn't happens to be just a change of signals. I was thinking that in this case the frictional force would be in static module, so it don't has value m.g.cos@.[coeficient]. Is that right?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
If you try the other direction, you will end up with a contradiction in your equations.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
i.e. sign of acceleration will not be the one you expect.
anonymous
  • anonymous
That's a question that my friend sent to me. I and him came to the same result for the oposite case, Which is a=-2.16666ms² , or something like this. We considered the value for the frictional force to be = m.g.cos@.[coeficient]. The expected would the answer to be a=-0.3444444...m/s², right? So my doubt is: what is wrong in this case? Is that frictional force isn't equal to m.g.cos@.[coeficient] , but instead < than it; or something else? And a p.s.: can both static and dinamic friction coeficients be equal to each other? (like in this question)
anonymous
  • anonymous
for the right answer: m1.g - T = m1.a T - m2.g.sen@ - m2.g.cos@.[coeficient] = m2.a for the confusion: m1.g - T = m1.a m2.g.sen@ - [frictional force] - T = m2.a Where T is the tension force
anonymous
  • anonymous
found a useful button here :D
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anonymous
  • anonymous
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Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
"can both static and dinamic friction coeficients be equal to each other?" It is not absolutely true, but it is often the case in problems. It makes them easier to solve.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Your calculations are correct, but the second case, with velocity down and acceleration up, can only be achieved if the crate was given an initial push downwards. Such a situation cannot exist if you start from rest.
anonymous
  • anonymous
I see :) Thanks for all! And sorry for taking so much of your time haha That's it. You really helped me Vicent.

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