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!

Reaction Force

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data
    The diagram shows the instant when a long slender bar of mass 4.9 kg and length 4.9 m is horizontal. At this instant the mass m= 6.5 kg has a vertical velocity of 4.0 m/s.

    If the pulley has negligible mass and all friction effects may be ignored, what is the magnitude of the reaction force on the bearing A
    problem_Moodle.gif

    2. The attempt at a solution
    So I split the reaction force into its x and y components Rx and Ry.

    For Mass m:
    T - mg = ma

    For the bar:
    Sum of the moments from A:
    let mass of bar = B
    length of bar = L
    B*g*L/2 - T.L = B*L^2/3*α

    Sum of forces in the x direction:
    Rx = B*a=B*ω^2*L/2

    Sum of forces in the y direction:
    Ry - Bg+T=B*(-a) = -B*α*L/2

    I solved and got Rx = 8N, however not sure how to find tension, a in the tangential, or alpha. I have tried simultaneous equations to find them but keep getting them wrong. Please help
     
  2. jcsd
  3. Sep 23, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, Jonski.

    You probably need to show your attempt at solving the simultaneous equations in order for us to find your error.

    Something to think about: Is the "a"in your equation T - mg = ma the same "a" as in your Rx and Ry equations?
     
  4. Sep 26, 2015 #3
    Hi TSny :smile:

    I get Rx = 32 N (towards right) and Ry = 37.46 N (downwards) . The magnitude of reaction force equal to 49.27 N .

    Are you getting the same result ?
     
  5. Sep 26, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hi Tanya.
    I get different values for Rx and Ry. For Rx, are you sure you're using the correct value for the speed of the center of mass of the rod?
     
  6. Sep 26, 2015 #5
    Okay .

    Rx = 8N (towards right) and Ry = 17.35 N (downwards) ?
     
  7. Sep 26, 2015 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK for Rx. But I still get a different answer for Ry.
     
  8. Sep 26, 2015 #7
    I will show you my working .
     
  9. Sep 26, 2015 #8
    Denoting mass of bar as M , length as L , mass of block as m ,

    For block of mass m , ##mg - T = ma ##

    Writing torque equation for the rod about the bearing, ##Mg\frac{L}{2} - TL = \frac{ML^2}{3}\alpha##

    ##\alpha = \frac{a}{L}##

    Solving this I get ##T = \frac{Mmg}{2(3m-M)}## and ##\alpha = \frac{3g}{2L} - \frac{3mg}{2L(3m-M)}## .

    Now writing torque equation for the rod about the CM , ## R_y\frac{L}{2} - T\frac{L}{2} = \frac{ML^2}{12}\alpha## .

    From this I get ##R_y = \frac{Mg}{4} + \frac{Mmg}{4(3m-M)}##
     
    Last edited: Sep 26, 2015
  10. Sep 26, 2015 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    So, positive value of ##a## implies what direction of acceleration for the block?

    So, positive value for ##\alpha## implies what direction of tangential acceleration for the left end of the rod?

    Does ##\alpha = \frac{a}{L}## or does ##\alpha = -\frac{a}{L}##?
     
  11. Sep 26, 2015 #10
    So,the only problem in post#8 is relationship between ##\alpha## and a ?

    ##\alpha = -\frac{a}{L}##
     
    Last edited: Sep 26, 2015
  12. Sep 26, 2015 #11

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes, your first two equations in #8 look fine. I have not checked your algebra for solving for T, etc.
     
  13. Sep 26, 2015 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Also, your equation for torque about CM looks good.
     
  14. Sep 26, 2015 #13
    Are you getting ##R_y = 28 N## ?

    Some doubts regarding why did we write ##\alpha = -\frac{a}{L}## ?

    It is because positive value of 'a' suggests block is going downwards which in turn means the rod is rotating clockwise . But since we have considered anticlockwise positive , we write ##a = -\alpha L## .

    Not sure , if this is the correct reasoning . You remember I once made an entire thread on sign issues. I am still doing some really sloppy work :olduhh: .

    How did you think about signs in this problem?
     
    Last edited: Sep 26, 2015
  15. Sep 26, 2015 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I got 28 N too. Your reasoning for why ##a## and ##\alpha## have opposite signs is the same as how I thought about it.
     
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: Reaction Force
  1. Reaction force (Replies: 2)

  2. Reaction force (Replies: 2)

  3. Reaction Forces (Replies: 6)

  4. Reaction Force (Replies: 2)

Loading...