1. Not finding help here? Sign up for a free 30min 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!

Help with this pulley question

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    An assembly shown in the picture, GH , HK, HNL are three cables. the mass of box L is 52Kg,
    a) determine the mass of box K in order to make angle theta as 25 degrees


    2. Relevant equations
    FX=0
    FY=0



    3. The attempt at a solution
    right i started with
    Fhn x cos(25) = Fhg x cos(0) as cos(o) is 1 i then wrote it out as
    Fhn x cos(25) = Fhg
    box L = 52Kg, i know HN = NL does the tension force NL= 52???
    i know that when i get HN by looking at the forces acting on L & using newtons 1st law to solve for NL i can calculate Fhg from the equation i started with. confused on how to get HN=NL thinks its 52.

    then at joint H, the vert comp of HN must equal the tension in HK, per newton 1st law in the y direction, and then i look at the forces acting on K to solve for its mass. i understand the theory i'm just confused about the calculations that need to be made, can anyone run through it with me please to find the mass of box K?????
     

    Attached Files:

  2. jcsd
  3. Apr 15, 2012 #2
    The 52kg block weighs 52x9.81N....this is the tension force in the string HN.
    If you look at point H can you see that the horizontal component of the tension = tension in string HG and the vertical component of the tension must be the weight of block K
     
  4. Apr 15, 2012 #3
    oh yes i forgot about the effect of gravity on the 52Kg block.
    ok i understand to a point about the mass of block k , so would the mass of block k be HN x (cos25+sin25) ? then divide by gravitional constant to get the Kg value ?
     
  5. Apr 15, 2012 #4
    You only need to consider the vertical component of the tension (510Sin25) to find the weight of block K
     
  6. Apr 15, 2012 #5
    oh yeah i see now why its only the vertical component, so just 510xsin25=215.5 that's all?
     
  7. Apr 15, 2012 #6
    That is the WEIGHT....I think you have to find the mass (easy?)
     
  8. Apr 15, 2012 #7
    yeah just divide by 9.81 i believe ?
     
  9. Apr 15, 2012 #8
    that's it !!!
     
  10. Apr 15, 2012 #9
    brilliant thanks for your help! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with this pulley question
  1. Pulley Question HELP (Replies: 2)

Loading...