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Homework Help: Newtonian System

  1. Aug 24, 2014 #1
    1. The problem statement, all variables and given/known data

    The system in the figure uses two masses m (I will call them m1 and m2) to raise a mass M. Consider that ropes are inextensible and that the mass of ropes and pulleys is zero. Find the acceleration of each body in terms of m,M,α and g. After that, indicate what is the minimum value of m so that M rise to a height yM in time t.

    Captura de pantalla 2014-08-24 a la(s) 10.57.32.png

    I will call the x component of acceleration of mass m1, m2, and M: am1(x), am2(x) and aM(x) respectively. Pulley A is the one in direct contact with M.
    3. The attempt at a solution

    My biggest issue was setting the reference system. I set x axes parallel to the floor and y axes positive downward. One constraint is that the length of the ropes are constant so that:
    length= ym1/sen(α) + 2yA + ym2/sen(α)
    length of shorter rope= yM-yA
    , where ym1, ym2 and yA are the coordinates of masses m1, m2 and pulley A in the y system. This leads to:

    Y called the tension on the longer and shorter rope T1 and T2 respectively. Because pulley A is massless, 2T1=T2 , because anything else would mean an infinite acceleration of the pulley.

    Newton's equation, using my reference system (N is the normal force perpendicular to the plane and is equal to m*g*cos(α):

    m1*am1(x)=-N*sen(α) + T1*cos(α)=(-m*g*sen(α)+T1)*cos(α)

    m1*am2(y)=-N*cos(α)+m*g-T1*sen(α)=m*g*sen2(α) - T1*sen(α)




    I solved this equations for the accelerations:

    aM(y)= (2*g*sen(α)) / (M/m - 2)

    am(y)= (-2g*sen2(α)) / (M/m - 2)

    am(x)= (2*g*sen(α)*cos(α) / (M/m - 2)

    For the second part of the problem I integrated aM(y) twiced, and because it is constant, it leads to yM=aM(Y)*(t2/2)

    I solved this equation for m:

    m=(yM*M) / (t2*g*senα + 2yM)

    Something must be wrong with this last equation, because m converges to 0,5 for a constant time when yM tends to infinity, but I can't see where is the problem, I supposed that there issomething wrong with the reference system I used.
  2. jcsd
  3. Aug 24, 2014 #2
    What I got for that is there, look at
    aM(y)= (2*g*sen(α)) / (M/m - 2)
  4. Aug 24, 2014 #3
    Let me ask you two questions:
    What reference system did you used?
    What about your kinematic constraints?
  5. Aug 24, 2014 #4
    Hi Vibhor, what do you think about my answer? :)
  6. Aug 24, 2014 #5
    Hello paalfis !

    I hope the wedges are fixed . Right ?
  7. Aug 24, 2014 #6
    I am sorry to ask this, but I don't want to mix my difficulties in english with the ones in physics :)
    What do you mean by wedges?
    If you mean the reference system, then yes, it is fixed for the entire system, that is why for equations concerning m1 and m2 there are a lot of trigonometric expressions there.
  8. Aug 24, 2014 #7
    Oh I must say something that I wrote wrong: The constraints I used are not ym/sen(α)=-yA=-yM but am(y)/sen(α)=-aA(y)=-aM(y)
  9. Aug 24, 2014 #8
    It is difficult to understand what you have done . But it doesn't look correct .

    No I meant the inclined surfaces under mass m .

    Assuming they are fixed , draw Free Body diagram of any one of m .You may consider coordinate axes along and perpendicular to the inclined surface .Write ∑F = ma1

    For M , again draw Free Body diagram .You may consider coordinate axes vertical and horizontal . For M you are only concerned with vertical direction.Write ∑F = Ma2 .

    The third thing you need is the constraint relation between a1 and a2 .

    Please take the symmetry of the system into consideration.
  10. Aug 24, 2014 #9
    I did those free body diagram but for x and y parallel and perpendicular to the floor respectively. I thought it was not right to compare a2 with a1 if these bodies are in different reference systems, that is why I tried to used one reference system for the three bodies.
  11. Aug 24, 2014 #10
    Please use three separate coordinate axes .You may align the 2-D coordinate axes for different blocks such that one axis lies along the direction of motion of respective blocks .

    For ex. the +x axis for left block will be downwards along the slope . The +x axis for right block will be downwards along the slope .The +x or +y axis for middle block will be upwards .
  12. Aug 24, 2014 #11
    Oh ok, thanks. I will do it that way, and refer to the x and y acceleration of each body with respect to its own coordinate axes.
  13. Aug 24, 2014 #12
    Ok, now my constraints are: aM(y)=am1(x)=am2(x) ; and again, 2T1=T2
    Newton equations are:

    solve for am(x):


    I was hoping to get m - M/2 instead of m+M/2, I checked it 10 times but that is my answer..
  14. Aug 24, 2014 #13
    It is quite difficult to understand your notations .Call the acceleration of m as a1 and that of M to be a2

    Fine .

    I hope you meant ##mgsin\alpha - T_1 = ma_1 ## . Right ?

    Not correct . Doesn't gravitational force act on M ?
  15. Aug 24, 2014 #14
    Oh.. silly mistake I am sorry, the problem is solved now, thank you very much
  16. Aug 24, 2014 #15
    Actually I am getting the same results as in my complicated first attemp to solve the problem, but I was forgetting the weight of M the whole time!! Now I feel a little better with my self :)
  17. Aug 24, 2014 #16
    Good :smile:
  18. Aug 24, 2014 #17


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    Just wondering, is the acceleration [itex]a=g\frac{M-msin(\alpha )}{M-m}[/itex] ?
  19. Aug 24, 2014 #18


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    It is not.

  20. Aug 24, 2014 #19
    My result:

    a=(g*(m*sen(a)-M/2)) / (m+M/2)
  21. Aug 24, 2014 #20
    Nathanael, your result does not makes sense if for the easy special case of sen(a)=1 and m=M/2
  22. Aug 24, 2014 #21


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    Yes, thank you. I made a careless mistake; I wrote [itex]Mg-T=Ma[/itex] instead of [itex]Mg-2T=Ma[/itex] :tongue: thanks
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