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!

Im stumped on this question please help!

  1. Oct 1, 2005 #1
    hey guys i am stumped on this question:

    mary dangles her watch from a thin piece of string, while burning rubber on her supercharged moped. she notices that the string makes an angle of 15degrees with respect to the vertical while the moped accelerates to maximum speed, which takes about 18s. what is the maximum speed of mary's moped?

    i am really stuck on this question and i need to show my work too. can anyone please help? thanks.
     
  2. jcsd
  3. Oct 1, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    What, exactly, have you done so far?
     
  4. Oct 1, 2005 #3

    Fermat

    User Avatar
    Homework Helper

    All you have to do is work out the acceleration on the watch/moped.
    The watch is at an angle of 15 deg.
    Draw a diagram of the forces on the watch. What are the only forces acting ?
    From this, you should be able to get the acceleration.
     
  5. Oct 1, 2005 #4
    i cant get anything, i just keep staring at the diagram but i cant figure it out, how do i figure out the acceleration by using the angle? do i brake it into the X and Y components? then what? :cry:
     
  6. Oct 1, 2005 #5
    is there no one that can help me?
     
  7. Oct 1, 2005 #6
    the only forces acting are the force of gravity (mg) and Ft (tension)
     
  8. Oct 1, 2005 #7

    Fermat

    User Avatar
    Homework Helper

    Have you got the diagram ?

    What forces have you got labelled on it ?
     
  9. Oct 1, 2005 #8

    Fermat

    User Avatar
    Homework Helper

    Ah, already answered.

    So, now resolve the Tension, Ft, into horizontal and vertical components.
     
  10. Oct 1, 2005 #9

    how? it only gives me the angle and the time, no forces or mass of the watch.
     
  11. Oct 1, 2005 #10

    Fermat

    User Avatar
    Homework Helper

    If you know the angle the tension is acting at, can't you just then resolve the tension, call it Ft, into horizontal and vertical components using that angle ?


    Take the mass of the watch as m.
     
    Last edited: Oct 1, 2005
  12. Oct 1, 2005 #11
    so this is what id be doing:

    Ftcos15= Ftx
    Ftsin15= Fty

    and then Ftx= m(ax)?

    then what do i substitute?

    the time comes from:
    accelerates to maximum speed, which takes about 18s. what is the maximum speed of mary's moped?

    the question gives it to me
     
    Last edited: Oct 1, 2005
  13. Oct 1, 2005 #12

    Fermat

    User Avatar
    Homework Helper

    Almost there.

    Ft.cos15 is actually the vertical component.

    You can equate this to the only other vertical force, the mass of the watch, hence you can get an expression for Ft.
     
  14. Oct 1, 2005 #13
    i always thought sine was the vertical componenet, when resolving vectors and i think thats true. also it doesnt give me the mass so how can i get an expression for Ft? can you please show me how?
     
  15. Oct 1, 2005 #14

    Fermat

    User Avatar
    Homework Helper

    I'll upload a sketch.

    The masss, m, should cancel out.,
     
  16. Oct 1, 2005 #15

    Fermat

    User Avatar
    Homework Helper

    See attachment.
     

    Attached Files:

    Last edited: Oct 2, 2005
  17. Oct 1, 2005 #16

    Fermat

    User Avatar
    Homework Helper

    What is your expression for Ft ?
     
  18. Oct 1, 2005 #17
    your attachment is pending approval, as for Ft i have no idea how to get an expression for it.
     
  19. Oct 1, 2005 #18

    Fermat

    User Avatar
    Homework Helper

    You had two eqns,

    Ftcos15= Ftx
    Ftsin15= Fty

    but they should have been,

    Ftsin15= Ftx
    Ftcos15= Fty

    (You will see why the cos and sin are the other way around when the attachment is approved. But it's just really because the angle, alpha, is the angle to the vertical. If it had been the aqngle to the horizontal, then sin would have been the vertical component.)

    The only vertical force is the mass of the watch, so Fty = mg. This gives,

    Ftcos15= Fty = mg
    ==============

    Now you have an expression for Ft, can you solve Ftx= m(ax) for a ?
     
  20. Oct 1, 2005 #19
    you have written an expression for FTy, but i need Ftx, is Ftx the same formula that being:
    Ftsin15=Ftx=m(ax)?
     
  21. Oct 2, 2005 #20

    Fermat

    User Avatar
    Homework Helper

    [tex]F_tsin15 = F_{t_x} = m(a_x)[/tex]
    [tex]F_tcos15= F_{t_y} = mg[/tex]

    [tex]F_tsin15 = m(a_x)[/tex]
    [tex]F_tcos15 = mg[/tex]

    Divide 1st eqn by 2nd eqn,

    [tex]\frac{sin15}{cos15} = \frac{a_x}{g}[/tex]

    [tex]a_x = gtan15[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?