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!

Rotational Equilibrium problem

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    (Figure is attached)
    A 1200.0 N uniform boom is supported by a cable,as
    shown in Figure 8-22.The boom is pivoted at the bot-
    tom,and a 2000.0 N weight hangs from its top.Find the
    force applied by the supporting cable and the compo-
    nents ofthe reaction force on the bottom ofthe boom.
    (See Sample Problem 8B.)

    Legend:
    t denotes torque
    T denotes tension
    d denotes distance from pivot
    Fg denotes gravitational force
    Rx denotes horizontal component of reaction force
    Ry denotes vertical component of reaction force.
    2. Relevant equations
    tnet = Fd sin theta
    Fnet =ma


    3. The attempt at a solution

    Take point A as pivot
    tnet = 0
    tnet = tT + tFg1 + tFg2 = 0

    +(T x 0.75L sin 90) - (Fg1 x 0.5L sin 25) - (Fg2 x L sin 25) = 0

    T = 1464 N

    Fnet = T + Fg1 + Fg2 + Rx + Ry = 0

    Fx = Fg1 cos 205 + Fg2 cos 205 + Rx = 0
    Rx = 2912 N

    Fg = Ry + Fg1 sin 205 + Fg2 sin 205 + T sin 90 = 0
    Ry = - 111

    R = Sqrt(-111^2 + 2912^2)
    R = 2915.7 N

    My angles for T, Fg1, and Fg2 are 90, 205, and 205, respectively. I got this wrong, and I'm curious why.
     

    Attached Files:

  2. jcsd
  3. Feb 3, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The weights act vertically.

    How are you using x and y?
     
  4. Feb 5, 2008 #3
    I took the forces perpendicular to the inclined boom, for torques ask for the forces to be perpendicular to the lever arm. I don't see why weights should act vertically, though. Shouldn't it be perpendicular to the boom, as I've done above?
     
  5. Feb 5, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Which way does gravity act?
     
  6. Feb 5, 2008 #5
    downwards, but doesn't it have a component that is perpendicular to the boom?
     
  7. Feb 5, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Sure.
     
  8. Feb 5, 2008 #7
    So in a sense, I am correct in taking the angles as they are above? I only ask this because my teacher took tension T at 155 degrees and F_g at 270 degrees, which doesn't make sense to me!
     
  9. Feb 5, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Your teacher is measuring angles with respect to the x-axis (horizontal), which is standard practice.
     
  10. Feb 5, 2008 #9
    We got different solutions though -- I wish it was just a matter of convention.
     
  11. Feb 5, 2008 #10

    Doc Al

    User Avatar

    Staff: Mentor

    When I get a few minutes, I'll look through your work. But since you use different x and y axes, could that be the problem? Aren't you supposed to give horizontal and vertical components?
     
  12. Feb 5, 2008 #11
    I think that is what I'd already done? Maybe it's a little messy up there -- it could definitely be redone more efficiently. For one thing, since tension is perpendicular, it's just simply positioned at 90 degrees to the boom. F_g would be in the opposite direction at 25 degrees, so the component would be F_g*sin(25). See, my reference is the boom -- that's where my horizontal is, and that is why, I think, our solutions differ.

    I'd appreciate it if you could look through my work!
     
  13. Feb 5, 2008 #12

    Doc Al

    User Avatar

    Staff: Mentor

    OK, I looked it over. Your work and angles look OK, given your choice of axis. (Except for some slight arithmetic differences.) Of course, if you don't translate your reaction force components into the form asked for (horizontal and vertical), then of course you will have different answers.

    (If someone's grading your work, you must make it easy for them to figure out what you're doing. You should state up front that your x and y axis are not horizontal and vertical, but parallel and perpendicular. But still, it's up to you to answer the question as asked.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rotational Equilibrium problem
Loading...