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!

Tension/Forces at Angles

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Fb is the spring balance which measures the total tension in the string. So T1 + T2 = Fb
    The system is at rest so a = 0.
    Weight W is hanging as shown in the diagram.
    T1 corresponds with angle α and T2 corresponds with angle β.

    ONLY THE HORIZONTAL FORCE CONTRIBUTES TO THE AMOUNT IN FB!

    Find the horizontal components of the force exerted by each string.

    2. Relevant equations
    T - mg = ma
    T1x + T2x = Fb


    3. The attempt at a solution
    I know that Force 1*cos(α) + Force 2*cos(β) = Fb

    Is Force 1 = Force 2 = W?
    Or is it Force 1 = Force 2 = 2W?
     

    Attached Files:

  2. jcsd
  3. Mar 2, 2013 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Forces on each side
    of ideal pulleys are the same. Draw free body diagrams of each pulley, starting with the lower one.
     
  4. Mar 2, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    First think about the string. If the tension varies along the string, what will the string do?
    Next, figure out what the tension is in the bottom part of the string.
     
  5. Mar 2, 2013 #4
    Upper pulley FBD = Tension to the left at angle α and I guess some type of support holding it up?

    Lower pulley FBD = Tension to the left at angle β as well as weight pointing down.

    For the lower one, it's supported by the Y component of that string, right?

    T*sin(β) = W
    T = W/sin(β)

    If forces on each side are the same, then (W/sin(β))*cos(α) + (W/sin(β))*cos(β) = Fb?
     
  6. Mar 2, 2013 #5

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    no. Although not shown , all pulleys are anchored down. If you look at the lower pulley, what is the tension in the section of the rope that is holding up the weight (draw an FBD of the weight).
     
  7. Mar 2, 2013 #6
    For the FBD, it's just tension pulling up and weight (mg) pulling down. T = W?
     
  8. Mar 2, 2013 #7

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  9. Mar 2, 2013 #8
    So W*cos(α) + W*cos(β) = Fb?
     
  10. Mar 2, 2013 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If W is the weight, yes. If W is the mass then use Wg.
     
  11. Mar 2, 2013 #10
    Thanks, everyone =)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted