1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Torque and Force of diving board

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    In the figure, the diver is simply standing at the end of the board, before beginning her diving action. The distance from support post B to the position of the diver's feet is L~=~ 2.1 m, and the distance from support post B to support post A is d~=~ 1.110 m. The weight of the diving board is 230 N and the mass of the diving board is uniformly distributed.

    http://session.masteringphysics.com/problemAsset/1000057181/10/diver.jpg

    Find the magnitude of the force exerted on the board by the supporting structure at point A.

    2. Relevant equations

    t = torque
    t = F*r

    3. The attempt at a solution

    For the first part of the question i had to find the torque on the front support B.

    t = F*r
    t = 58(9.8)(2.1) = 1193.6 Nm

    From here i thought to get the force on point A i could

    find the torque on point A and subtract the torque at point B

    ta = 58(9.8)(2.1+1.11) = 1824.56

    ta = 1824.56 - 1193.6 = 630.96

    then to solve for the force on point a

    Fa = ta/(2.1+1.11)
    Fa = 630.96/(2.1+1.11) = 196.56 N

    Does this look right? I feel it is wrong.

    Thank you for any help :)
     
  2. jcsd
  3. Apr 6, 2010 #2
    I'm sort of confused about what you're doing. What exactly was the question for the first part? If it asked for net torque about support B, what you've done isn't right. You've only taken in to account the torque on support B due to the person.

    I ask because the answer to the first part needs to be right to get the second part.
     
  4. Apr 6, 2010 #3
    The question for the first part that i showed my work for was

    Calculate the magnitude of the torque about the front support post (B) of a diving board

    and i got this part right. = 1193.6 Nm


    The question i am looking for now and i cannot find is

    Find the magnitude of the force exerted on the board by the supporting structure at point A.

    Thanks for the help
     
  5. Apr 6, 2010 #4
    Okay, I think I might have figured this out. I'm still confused about how 1193.6 Nm can be the answer to the first part, unless the question asked for the torque about point B due to the diver.

    Also, the diver's mass is 58 kg, right? I'm inferring this from your first post.

    So to figure out the force at A, you need to calculate the NET torque about point B. Since the situation is in static equilibrium, it's not moving. Hint 1: What does the torque have to be equal to in that situation?

    Hint 2: Start by calculating the total torque about B. Make sure you include the forces from A, the weight of the board, and weight of the person in this calculation.
     
  6. Apr 6, 2010 #5
    Hint 1: wouldnt it just equal zero since its in static equilibrium?

    Hint 2:

    I am a little confused on this part.

    sorry my book does not seem to cover this section very well so i am just a little lost, yet my professor is pushing us with a ton of hw on it.

    am i even close when i write this?

    t_b = (F_a+58*9.8+230)*(2.1)
     
  7. Apr 6, 2010 #6
    Yep, it would be equal to zero.

    t_b meaning torque about b? I'm not sure what you did to get that answer, but it doesn't look right to me.

    Let's just start by calculating the torque about support B from the beginning. Whenever you calculate a torque, you define a pivot point. In this case it's the support B. Then you calculate the individual torques from all the forces acting on the board about the pivot point, and add them together. This gives you the net torque.

    To find the torque from just one force, multiply the magnitude of the force by its distance to the pivot point.

    So, finding the torque about point B due to the person gives you t = (58 kg)(9.8 m/s^2)(2.1 m).

    Now you need to do the same for the mass of the board and the force from support A. When you find them individually, you add em up. So try that and let me know what you get.
     
  8. Apr 6, 2010 #7
    Oh ok i think i get it now. that was a good explanation thank you.

    ok so the torque at b due to the person like you said is

    t = 58(9.8)(2.1)

    which was the first part i had to calculate.

    now for the torque at B due to the board and point A i get

    t = (230+58*9.8)(1.11)

    so the total torque at B would b

    t_btotal = (230+58*9.8)(1.11) + 58(9.8)(2.1)

    does that look right?

    thanks again for the help. i really appreciate it.
     
  9. Apr 6, 2010 #8
    Okay, almost!

    You shouldn't be getting a 58 in the torque due to B and due to the board's weight. By the way, it would help if you wrote in your units, it just makes things clearer :)

    The torque due to point A is f_a (force due to A) multiplied by the distance to B. Which is 1.11 m, looks like you got that. But torque due to point A is simply f_a*0.11 m.

    Now, torque due to the weight of the board is a little more complicated. The weight of the board acts at its center of mass. Since the board is uniformly distributed, this is just the middle of the board. So, think of it like a point force acting at the middle of the board. What's the torque due to the weight of the board only? (Think about what the distance from the middle of the board to B is...)
     
    Last edited: Apr 6, 2010
  10. Apr 6, 2010 #9
    i edited mine a little. i added in the total torque at B (t_btotal)
     
  11. Apr 6, 2010 #10
    Well, you have the right idea, but there's a problem with the first term of t_btotal. You'll see what I mean when we get to the next step, once you do the torque from the weight of the board.
     
  12. Apr 6, 2010 #11
    torque due to the board would just be

    t = 230N(.495m)

    i got the .495 from this ----- 2.1m+1.11m = 3.21m

    3.21m/2 = 1.605m

    1.605m is the center of mass of the board so the distance from this to the point B is 0.495 m
     
  13. Apr 6, 2010 #12
    Exactly! :)

    So to summarize, we have 3 individual torques:
    torque due to the person = (58 kg)(9.8 m/s^2)(2.1 m)
    torque due to support A = f_a(1.11 m)
    torque due to the board = (230 N)(0.495 m)

    Of course, the force from support B doesn't exert a torque because it acts at the pivot point.

    So now we need to add them together to get the net torque. Makes sense up to here, right?

    There's a catch when you add them up, though. Some torques are going to be negative, some will be positive. If the torque would rotate the board counterclockwise, it's positive. Otherwise, negative. Think especially carefully about force A.
     
    Last edited: Apr 6, 2010
  14. Apr 6, 2010 #13
    great ok so this is what i get when adding them up

    (58kg)(9.8 m/s^2)(2.1m) + 230N(.495 m) - f_a(1.11 m) = 0

    does this look right? can i just solve it for f_a now?
     
  15. Apr 6, 2010 #14
    do i have some signs wrong?

    when i think about it again i am thinking it is

    -(58kg)(9.8 m/s^2)(2.1m) - 230N(.495 m) + f_a(1.11m) = 0

    which i guess is the same all in all but i want to make sure i do it all right instead of just some of it right :)
     
  16. Apr 6, 2010 #15
    Yes, what you did second is right.
     
  17. Apr 6, 2010 #16
    thank you that was all correct :)

    i am having some troubles with the next part. it asks

    Find the magnitude of the force exerted on the board by the supporting structure at point B.

    this is what i am thinking

    torque due to a - torque due to the person + f_b(0) = 0

    so i thought it would just be zero but that was wrong :(
     
  18. Apr 6, 2010 #17
    First, think about what you're taking the torque about. You always need to pick a pivot point. So what pivot point are you using this time?
     
  19. Apr 6, 2010 #18
    i think i should use A for my pivot point.

    so from that i am getting

    -230N(1.605m) - (58kg)(9.8 m/s^2)(2.1m+1.11m) - f_b(1.11m) = 0
     
  20. Apr 6, 2010 #19
    Close!

    You're missing a distance in your second term, and you have some sign problems.
     
  21. Apr 6, 2010 #20
    oopps

    i edited my signs and added in the distance i forgot lol

    does that look right?

    thanks again for all the help :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook