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!

Force & Acceleration of a Ridgid Body

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    See attachment

    The door has a weight of 200lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12ft to the right in 5 sec. starting from rest. Also, find the vertical reactions at the rollers A and B.


    2. Relevant equations
    F=ma


    3. The attempt at a solution

    200lb --> m = 6.21619 slug

    F(x) = 6.21619a
    F(y) = N(ab) - 6.21619(32.2) = 0, N(ab) = 200.161318 N

    Moment @ G = F(2ft) - N(a)(6ft) + N(b)(6ft) = 0

    N(a) + N(b) = 200.161318

    Not sure where to go from here
     

    Attached Files:

  2. jcsd
  3. Apr 30, 2009 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You forgot that the acceleration can be solved by the basic kinematic equation of motion from the givens: the box moves 12 feet in 5 seconds. Solve for the aceleration, then the rest looks OK (except please get rid of those decimals after the 200!).
     
  4. Apr 30, 2009 #3
    Ok not sure which one to use. I'm confused about my variables.


    heres what i tried

    v = v(o) + at where v = 12/5, v(o) = 0, and t = 5, a = .48

    s = s(o) + v(o)t + .5at where s = 12ft, s(o) = 0, v(o) = 0, t = 5, a = 4.8

    v^2 = v(o)^2 + 2a(s - s(o)) where v = 12ft/5sec, v(o) = 0, s = 12ft, s(o) = 0, a = .24

    Whats up with that

    Also why not use the decimals. it is the force of gravity, ma
     
  5. May 1, 2009 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, 12/5 is the average velocity, not the instantaneous velocity
    this is the right equation to use, but you forgot to square the 't' as noted in red , so you got the wrong answer
    nahh, v is not 12/5
    The force of gravity, which is the objects weight, is given as 200 pounds. You complicated it .
     
  6. May 1, 2009 #5
    ok

    so my a = .96 ft/s^2

    so then F = ma where m = 6.21619 slug and a = .96 ft/s^2
    F = 5.97 slug ft/s^2

    & the Vertical reaction on the rollers is 100lb each.

    It does not add up though when u apply it to the moment at G
     
  7. May 1, 2009 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK
    OK, but a slug ft/s^2 is called a 'pound' for short. F=5.97 pounds (call it 6 pounds)
    No, that's not right. They are not equal. All you know right now is that the sum of the forces of each roller on the block must equal 200 pounds. Don't assume they are equal.
    They should after you do the moment thing. I suggest, however, that you sum moments about one of the rollers instead of about G. That way, life will be a lot easier, because you'll be able to solve for one of the roller forces immediately, and the other will then come out of the previous 'sum of F_y = 0' equation.
     
  8. May 1, 2009 #7
    Ok so i will get the moment about rolller a

    5.97(9) + 12*F(b) - 200(6) = 0

    F(b) = 95.5225
    F(a) = 104.4775
     
  9. May 1, 2009 #8

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Look good! But don't forget your units, and you should round off your values. I would say that F_a = 104 pounds (lb. or lbs.) and F_b = 96 lbs. And F = 6 lbs.
     
  10. May 1, 2009 #9
    thanks for your help working through this sucker with me
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook