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Maximum weight of block m2 for which block m1 will remain at rest

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    So this was part of a two part thing, the first part was just to find out which statements were true and i figured that out but as far as this i just dont know what to do. this first stuff is for the problem that i already have and it is the second stuff that i need the help with.
    a block m1 sits on a table. There is static friction between block and table. Block m2 hangs from a knot, as shown. Call the tension in the rope connecting the knot to m1, "T1". Call the tension in the rope connecting the knot to m2, "T2". Call the tension in the third rope (the one tipped up by an angle theta, connecting the knot to the wall), "T3". The system is in equilibrium.
    so this next stuff is what i need the help on
    block m1 weighs 851 N. The coefficient of static friction between the block and the table is 0.24 and the angle theta is 27.6o. Find the maximum weight of block m2 for which block m1 will remain at rest
  2. jcsd
  3. Oct 6, 2008 #2
    It would help if we could see a diagram thanks
  4. Oct 6, 2008 #3
  5. Oct 7, 2008 #4
    Okay so lets have a look at this. We know the system is at rest, that means that the sum of forces = 0.

    All forces in the x = 0 and all forces in y = 0 is a simpler way of looking at it.

    So in the x plane we have T1, friction and x component of T3.
    in the y plane we have gravity, normal force, T2 and y componenet of T3.

    Write out both of these equations equalling to zero and solve.
  6. Oct 7, 2008 #5
    how would you use all of those things to come up with one equation though. so like i know that for a tension with something on a table T1=m1a+m1g and that T2=m2g-m2a. but i dont know about getting any of the other stuff
  7. Oct 7, 2008 #6
    Well it's very simple, turn the worded equations I wrote into algebraic equations:

  8. Oct 7, 2008 #7
    so it would be Ffric=T1*T3?
    and then N=9.8(T2*T3)
  9. Oct 7, 2008 #8
    Whoa slow, down, we have to sum up the forces in each plane (the forces are the ones I mentioned in my quote) and let them equal zero.

    Make sure to add them, not multiply.

    so, for the x plane (we'll define the right as positive); T3cos(theta) + T1 - Friction = 0
    for the y plane (we'll define up as positive); T3sin(theta) + T2 - mg = 0
  10. Oct 7, 2008 #9
    ok and then when we are trying to find what T1, T2, and T3 equals do we use T1=m1a+m1g and that T2=m2g-m2a?
  11. Oct 7, 2008 #10
    What would you use as 'a'?
  12. Oct 7, 2008 #11
    a/g=m2/(m1+m2) so a=g(m2/(m1+m2))
  13. Oct 7, 2008 #12
    What I am saying is; what is 'a'? What is it caused by and what does it affect?
    Consider this before you answer, nothing in the system is moving.
  14. Oct 7, 2008 #13
    so then a would equal 0 and T1=m1g and T2=m2g?
  15. Oct 7, 2008 #14
    oh but since i dont know what m2 is i wouldnt be able to use this to find T2. and also how do i find out what T3 is?
  16. Oct 7, 2008 #15
    Sorry you did not answer my question of what you are saying a is. I'm not asking for a mathematical representation of it, I was wondering what you thought caused a.

    I can tell you right now that T1 does not = m1g and T2 does not equal m2g.

    There are other forces acting. You can see that T1 is going to be the x component of T3 opposing the friction of m1. Right there it tells you that T1 = m1g*friction + T3cos(theta)

    T2 = m2g + T3sin(theta).

    Please don't try to answer this question until you understand why this is.
    All I am doing is summing up the forces in x and y
  17. Oct 7, 2008 #16
    ok. i still just dont understand all of what you are telling me. i am not sure really by what you mean when you say x component of T3 opposing the friction of m1
  18. Oct 7, 2008 #17
    Ok, just to make sure, do you understand that because T3 is on an angle, we can turn it into two imaginary forces, one along the x axis, and one along the y. When you add these forces together you will get T3.

    Now that we have got that, the x component of T3 is pulling towards the right, and subsequently tries to pull T1, as this happens, friction is going to act. However friction is going to act in the opposite direction to the force pulling the block (so it becomes harder to pull).

    Therefore the x component of T3 is in the opposite direction to friction caused by m1.
  19. Oct 7, 2008 #18
    sorry man i am still very lost. i still do not understand what is meant by x and y component. i remember my teacher talking about it during class but cant recall exactly what it is and how it affects anything
  20. Oct 7, 2008 #19
    Okay, lets say that we have a right angled triangle sitting on a graph, where one side is along the x axis, another side is along the y axis, and the hypotenuse is the line that joins the two ends.

    You can see that the x axis and the y axis combined create the hypotenuse in a relation ship of h2 = x2 + y2

    In that same way, if the hypotenuse the direction of a force, it can be broken up into two (or many more) components, that although are not new forces, combine to make the original force we were given.

    So we have broken T3 which you can see is not along either x or y axes exclusively into two forces that are along x and y axes.

    Using trigonometry, from image you have provided, T3x = T3cos(angle)
    and T3y = T3sin(angle)

    Now all of the forces in the system are only either in the x plane or the y plane.

    We know the system is not moving, therefore all forces in the x plane sum up to 0, as do the forces in the y plane.
    Does that make more sense?
  21. Oct 7, 2008 #20
    so i was actually able to get the answer. what i did was multiply the weight in N times (tan theta)(coefficient of friction) and that gave me the answer. thanks for all of ur help tonight though man
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