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Three masses and Two pulley

  1. Nov 25, 2015 #1
    1. The problem statement, all variables and given/known data
    I have 3 masses with 2 pulleys which is in static equilibrium.
    M1 & M3= 50.8kg
    M2 = 90kg
    Gravity that I used = 10m/s²
    Part 1
    Prove that the system is in static equillibrium

    Part 2
    If I were to add another 10kg of mass to M3 is the force exerted from M3 the same force as the tension from the rope. Assuming the pulley and rope is massless?
    And how do I find the displacement of all the masses (M1, M2, M3)?
    Will the system be in an equilibrium or not?
    If it's not, show how do you find the acceleration, velocity and displacement for all three masses.

    2. Relevant equations
    F = ma


    3. The attempt at a solution
    For part 1, from the equation I have drawn a FBD for M2 and have calculated the net forces of F(x) and F(y) for M2 which is:
    Fx = 508 * cos (62.5) - 508 * cos (62.5) = 0;
    Fy = (508 * sin (62.5) + 508 * sin (62.5)) - 900 = 0 (I round it off due to calculation approximation);

    Am I doing the part 1 correctly? As I am assuming that the forces from M1 & M3) is acting on M2 so I took the forces from there and apply it on the FBD for M2.

    For part 2, I can't really continue if I'm not sure whether part 1 is right (correct me if I'm wrong whether can I still continue on part 2). Even then, I'm trying to find the relationship between F = ma and the a, v, and s(displacement) cause I have been searching for the equation but I seem to be more lost than ever.

    Thanks once again and I hope I don't sound too dumb as I am new to statics.
     

    Attached Files:

    Last edited: Nov 25, 2015
  2. jcsd
  3. Nov 25, 2015 #2
    Part 1 seems to be correct.

    For part 2: Think about what is the necessary condition for a second equilibrium after increasing the mass M3 (hint: the angles α and β will be different then).
     
  4. Nov 25, 2015 #3
    For part 2, the necessary condition for an equilibrium on the system is that all net forces must equal to 0, however, when I calculated after adding 10kg on M3, wouldn't the Fx and Fy for M2 also changes (correct me if I'm wrong), which means M2 will travel along the rope of M3, which will pull M1 up, resulting in an infinite acceleration thus can't be in an equilibrium? So sorry if I sound dumb or stupid, am really new at this and have been self taught from youtube tutorials and etc.
     
  5. Nov 25, 2015 #4

    haruspex

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    I'm not sure what is being asked in part 2. Are you to find a position of the masses which is at equilibrium (or show there is none such), or find the instantaneous acceleration from the given position?
     
  6. Nov 25, 2015 #5
    Sorry for not being clear, it is to find a position of the masses after the addition of 10kg at M3 at equilibrium. However I am just curious of the instantaneous acceleration that will occur during the addition of 10kg mass (thus the displacement/velocity/acceleration). That's why I am kind of stuck here (this question was given by my lecturer to explore actually. :)
     
  7. Nov 25, 2015 #6

    haruspex

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    Ok. It was this line that threw me:
    The acceleration etc. will depend on the current position. If there's no new equilibrium, there's no position defined. And if you are going to consider accelerations etc. based on the staring point then it won't stop moving when at the equilibrium point.

    Anyway, let's first try to find a new equlibrium. Have you drawn a free body diagram? Define some unknowns that describe the arrangement and the forces in it. What equations can you write down?
     
  8. Nov 25, 2015 #7
    Correct

    That's true and as haruspex wrote this only applies for the first moment - then the direction of the net force changes.

    Don't be to strict with yourself - it needs a little bit of training to solve this kind of problems. For the second equilibrium: Start to find the condition in horizontal direction.
     
  9. Nov 25, 2015 #8
    To move to the new equilibrium, which way do you think the 90 kg mass has to move? Would the other masses move up or down? How would that affect the geometry?

    Chet
     
  10. Nov 25, 2015 #9
    I can only imagine that the affected mass (M3) will go down, thus M2 will go north east due to the resultant force applied from M3, which then M1 will have to move up to accommodate to the movement of M2.

    Here's what I got so far:

    Fx = 608 * cos (62.5) - 508 * cos (62.5) = 234.6 (rounded up);
    Fy = (608 * sin (62.5) + 508 * sin (62.5)) - 900 = 89.9 (I round it off due to calculation approximation);
    Thus, the Resultant force is: R = √Fx2 + Fy2 = 251.2;
    and the resultant angle is: Rangle = arctan(89.9 / 234.6) = 21;

    I'm not sure whether I'm doing it correctly but then again, after finding this, is the system at equilibrium when the mass with 90kg moves over to its new position?
     
  11. Nov 25, 2015 #10
    The angles will change also.
     
  12. Nov 25, 2015 #11
    Meaning? Which angles? is it the α and β? I think my train of thought is stuck as I really do not know how to find the new α and β. So my calculations was wrong? No wonder!
     
  13. Nov 25, 2015 #12
    Yes. α and β both change. You need to find the new α and β. You do this by writing the force balances in the x and y directions in terms of α and β (algebraically), and solving for the new α and β.
     
  14. Nov 25, 2015 #13
    No, the new unbalanced force will cause a finite acceleration. First, calculate the new equilibrium angles (at least show the equations; solving them is tricky).
     
  15. Nov 25, 2015 #14
    So it will be something like,
    Fx = 608 * cos (α) - 508 * cos (β) = 0;
    Fy = (608 * sin (α) + 508 * sin (β)) - 900 = 0

    I then used algebra to solve for Fx and got
    (α) = cos-1(508 cos (β) / 608) = cos-1(0.84 cos (β)

    This part I am stuck as it got a little tricky honestly.
    does cos-1 cancel out cos (β)?
     
  16. Nov 25, 2015 #15
    Why not use Cosine Law.
     
  17. Nov 25, 2015 #16
    Solving these two simultaneous equations for the two angles is a little challenging mathematically, but doable. If you do it by brute force, though, it will be pretty difficult.
     
  18. Nov 25, 2015 #17
    I have M1 and alpha on the left, and M3 and beta on the right on your diagram.
    In my equations, alpha and beta are reversed from yours(?).
    Anyway, I just used trial and error and got 3 significant figures in 7 trials.
     
  19. Nov 25, 2015 #18
    Yes apologies it is supposed to be
    Fx = 508 * cos (α) - 608 * cos (β) = 0;
    Fy = (508 * sin (α) + 608 * sin (β)) - 900 = 0

    (α) = cos-1(608 cos (β) / 508) = cos-1(1.2 cos (β))
    replace (a) to Fy's equation I get:
    508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) - 900 = 0;
    508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) = 900;

    I wanted to factorise but it doesnt seem that there's any common values.
    Anyway,
    @azizlwl cosine law? as far as I can remember I think need to know the three sides which is currently unknown at the moment, no?
     
  20. Nov 25, 2015 #19
    $$\cosα=\frac{508}{608}\cosβ$$
    $$\sinα=\sqrt{1-\left(\frac{508}{608}\right)^2\cos^2β}=\sqrt{1-\left(\frac{508}{608}\right)^2+\left(\frac{508}{608}\right)^2\sin^2β}$$
    Substitution in Fy equation yields:
    $$\sqrt{608^2-508^2+508^2\sin^2β}+508\sinβ=900\tag{1}$$If we multiply numerator and denominator of the left hand side of this equation by##\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ##, we obtain:
    $$\frac{608^2-508^2}{\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ}=900\tag{2}$$
    If we invert Eqn. 2, we obtain:
    $$\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ=124\tag{3}$$
    If we subtract Eqn. 3 from Eqn. 1, we obtain:
    $$1016\sinβ=776$$or$$\sinβ=\frac{776}{1016}$$So, β=49.8 degrees

    Chet
     
  21. Nov 26, 2015 #20
    You have 3 sides(vectors) 900, 608 and 508 in equilibrium.
     
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