1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pulley blocks problem

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Q. In the figure shown, find out the value of θ [ assume string to be tight ]

    2. Relevant equations



    3. The attempt at a solution

    If the pulley was not moving ,then we could have equated the component of velocities of the two blocks along the string .

    But,since the pulley is moving,something extra needs to be done .Not sure what .

    I would be grateful if somebody could help me with the problem .
     

    Attached Files:

  2. jcsd
  3. Mar 16, 2014 #2
    I am not sure but how about working in the frame of pulley?
     
  4. Mar 16, 2014 #3
    Could you elaborate what do you mean by working in the pulley frame ? What benefit does it give in the context of the problem ?
     
  5. Mar 16, 2014 #4
    I meant that if you work in the frame fixed to the pulley, the blocks have an additional component of velocity downwards. You can then equate the velocity along the string. (And that gives one of the answers mentioned.)
     
  6. Mar 16, 2014 #5
    I analyzed this a little differently. Drop a normal from the pulley to the line joining the tops of blocks A and B. Call the length of this normal h. You then have two right triangles. Express the other sides of these two right triangles trigonometrically in terms of h (using trig functions of 30 degrees and θ). Let a small increment of time Δt pass. Express then new sides of the two triangles in terms of Δt. The sum of the hypotenuses of these new triangles must be the length of the string, which is constrained to be unchanged. So set the sum of the new hypotenuses equal the sum of the original hypotenuses. Expand each of the two hypotenuses as a linear function of Δt. Certain terms will cancel from both sides of the equation. Set the coefficient of Δt equal to zero. Solve for θ, or substitute each of the possible answers into the equation, and see which one satisfies it.

    Chet
     
  7. Mar 20, 2014 #6
    Hi Chet...

    Sorry for the late response .

    Let h be the length of the normal ; x length of the left string and y length of the right string ; a and b be the length of the bases of left and right triangles respectively at t=0.

    x=hcosecθ and y=hcosec30°
    a=hsecθ and b=hsec30°

    Now after time Δt ,

    a' = a-3.25Δt
    h' = h+Δt
    b' = b+√3t

    Now x+y = √(a'2+h'2) + √(b'2+h'2)

    [tex]hcosecθ+hcosec30° = \sqrt{(hsecθ-3.25Δt)^2+(h+Δt)^2} + \sqrt{(hsec30°+\sqrt{3}Δt)^2+(h+Δt)^2} [/tex]

    Is it correct ?
     
    Last edited: Mar 20, 2014
  8. Mar 20, 2014 #7
    Ooops. Those sec's should be cots.

    Chet
     
  9. Mar 20, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You know the motion of everything. The blocks move horizontally, with the given velocities.
    The pulley moves vertically upward with 1 m/s.

    The string is tight, and it can move on the pulley. The total length is constant.

    Choose the origin on the ground, just below the pulley. The positions of the blocks are A (xa,0) and B(xB,0). The pulley is at P(0,y). Write up the length of the pieces LA, LB. LA+ LB=constant. Take the derivative with respect to time. You know the derivative of dxA/dt=3.25, dXB/dt=√3, dy/dt=1. y/(-xA)=tanθ, y/xB=tan(30°)...


    ehild
     
  10. Mar 20, 2014 #9
    Yes...poor effort on my part :redface:.

    If I open the expression under the radical sign ,I get lot of terms none of which seem to cancel.

    Now what to do with terms containing Δt and Δt2 ?

    Sorry...I didn't quite understand this .
     
  11. Mar 20, 2014 #10
    Discard the (Δt)2 terms.

    [tex] \sqrt{(hcotθ-3.25Δt)^2+(h+Δt)^2}≈\sqrt{h^2(cot^2θ+1)+2hΔt(1-3.25cotθ)}≈hcscθ\sqrt{1+\frac{2hΔt(1-3.25cotθ)}{h^2csc^2θ}} [/tex]
    [tex] \sqrt{(hcotθ-3.25Δt)^2+(h+Δt)^2}≈hcscθ+Δt(sinθ-3.25cosθ) [/tex]

    Chet
     
    Last edited: Mar 20, 2014
  12. Mar 20, 2014 #11
    Thanks ehild :smile:...Very nice and systematic way of looking at the problem.

    But I seem to be struggling with sign errors :shy:. I don't get correct answer with the data you have given .Yet it looks fine to me.

    If instead of treating (xa,0) and B(xB,0) as coordinates of A and B ,I look them as length (distance ) from the origin ,I get correct answer .

    Then we would have

    dxA/dt = -3.25, dXB/dt=√3, dy/dt=1. y/(xA)=tanθ, y/xB=tan(30°)

    But I don't understand what is the problem with treating xa and xB as coordinates instead of lengths(distances).
     
    Last edited: Mar 20, 2014
  13. Mar 20, 2014 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I can not know if you do not show your work....

    ehild
     
  14. Mar 20, 2014 #13

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I like Pranav's approach to this problem. No calculus or expressions for the length of the string needed. In his frame of reference, the pulley has only rotational motion.
     
  15. Mar 20, 2014 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, it is very nice...

    ehild
     
  16. Mar 20, 2014 #15

    utkarshakash

    User Avatar
    Gold Member

    This problem can be simplified by using Virtual-Work Method.
     
  17. Mar 20, 2014 #16
    Please see this post in conjunction with my earlier post.

    [tex]\sqrt{{x_A}^2+y^2} + \sqrt{{x_B}^2+y^2} = constant[/tex]

    [tex]\frac{2x_A\frac{dx_A}{dt}+2y\frac{dy}{dt}}{2\sqrt{{x_A}^2+y^2}} + \frac{2x_B\frac{dx_B}{dt}+2y\frac{dy}{dt}}{2\sqrt{{x_B}^2+y^2}} [/tex]

    [tex]\frac{\frac{dx_A}{dt}+\frac{y}{x_A}\frac{dy}{dt}}{\sqrt{1+{\frac{y}{x_A}}^2}} + \frac{\frac{dx_B}{dt}+\frac{y}{x_B}\frac{dy}{dt}}{\sqrt{1+{\frac{y}{x_B}}^2}} [/tex]

    On putting in values given in post#8 we get

    [tex]\frac{3.25-tanθ}{secθ} + 2 = 0 [/tex] which gives incorrect result .

    Whereas if I use values in post#11 I get [itex]\frac{-3.25+tanθ}{secθ} + 2 = 0[/itex] ,which gives correct result .

    Could you reflect on this .
     
  18. Mar 20, 2014 #17

    utkarshakash

    User Avatar
    Gold Member

    I think the problem lies in assuming the coordinate of left block as (x_a,0). It should be (-x_a,0) instead, where x_a>0. Now, if you closely look at [itex] d x_a /dt [/itex] which is the rate of change of x_a, you will find that it is -ve. This is because as the left block progresses towards the origin, x_a(which was initially some +ve quantity) approaches zero which clearly shows that x_a is decreasing and a -ve sign must be placed along with 3.25. Thus the second equation is the correct one.
     
  19. Mar 20, 2014 #18
    Your reasoning looks quite convincing . Thanks :)
     
  20. Mar 20, 2014 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You pulled out xA from the square root, which is negative. It should have been |xA|. ##\sqrt{(-2)^2}=2## and not -2.
    I pulled out y and got correct result.

    ehild
     
    Last edited: Mar 20, 2014
  21. Mar 20, 2014 #20
    Excellent troubleshooting !!!

    If you had given me one whole day , i wouldn't have found that error :tongue:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pulley blocks problem
Loading...