Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Statics Problem: Weight Hanging from a Rope

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Mass hanging from the pulley in 50kg, the mass on the rope is 1kg. Please see attachment for details. The answer will be in terms on L.


    2. Relevant equations
    The sum of the forces in the y-direction and x-direction are equal to 0.


    3. The attempt at a solution
    Adding the moments at the pulley and equating them to 0; we have that T=50.
    Now a free body diagram of the second weight. 2*50sin(x)=1, sinx=(1/100)=H/X
    X=(L^2+H^2)^(1/2), substituding this in (1/100)=H/X, and solving for H, I find H=L/100.
    Which is not the answer I have in the solutions.
    Any help will be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Jul 20, 2010 #2
    Can you explain to me "2*50sin(x)=1"?
     
  4. Jul 20, 2010 #3
    When I draw the FBD of the 1kg weight; I addthe forces in the y-direction
    50*sin(x)+50*sin(x)=1
     
  5. Jul 20, 2010 #4
    What is the answer that they give?
     
  6. Jul 20, 2010 #5
    I have attached the solution. I cannot follow what they are doing; if anybody does please help.
     

    Attached Files:

  7. Jul 20, 2010 #6
    The key here is that they are asking how much the 50kg weight will move up due to the 1kg weight. the Solution shows the change in length of the rope over the horizontal section, which is the same as the change in height of the vertical section, which is the same as the 50kg mass.
     
  8. Jul 20, 2010 #7
    Ok, I understand that now. But I still don't get where the 50/L is coming from.
     
  9. Jul 20, 2010 #8
    sin(tetha)=50/L
     
  10. Jul 20, 2010 #9
    Ok got it, thanks for the explanation.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook