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!

Velocity from F(t) graph

  1. Oct 12, 2013 #1
    We are pulling a box with a mass = 8kg up a slope (ψ=30°). Force of a hand is given F(t)=F0*sin(Ωt). F0= 150N, Ω=(2*∏)/t0, t0=30s. After tf=15s we don't pull it anymore.
    a) Find average velocity at [t=12s to t=10s], current velocity at t=11s and v(t) before and after the end of pulling.
    b) Find x(t).

    I am having problems to find velocity ot of force function. Can anyone help me please? :)
    Thank you in advance. :)
     
  2. jcsd
  3. Oct 12, 2013 #2
    show your work.
     
  4. Oct 12, 2013 #3
    Ok. First I calculated Fd (the component of Fg down the slope => sin30°*Fg=40N). Then I calculated the F(t=10s)=129,9N and F(t=12s)=88,17N. Then I figured out that the area under the graph F(t) = integral = F*Δt=m*Δv. So F*Δt= 221,32Ns. Then I divided by 2s (Δt) so that I got Force alone on one side. I got 110,66N. Now, I substracted Fd from this F and got 70,66N=(m*Δv)/Δt. From that equation I got Δv=17,67 m/s, but from now on I couldn't find velocities out of the Δv.
    I feel terrible, because I stared in this exercise for about 4 hours and I tried everything, always got different results.
     
  5. Oct 12, 2013 #4
    Hi gasar8

    Welcome to Physicsforums!!!

    The force by hand is a sinusoidally varying force given by F=150Sin(2πt/30).

    The net force along the incline is F - mgsin30°

    So, 150Sin(2πt/30) - 40 = 8dv/dt

    Integrate with proper limits and you will get velocity as a function of time v=f(t) .From this you can get velocity at 11s and velocity before pulling.Similarly you can get an expression for velocity after pulling stops.

    Again write v= dx/dt = f(t) .Integrate with proper limits and you have displacement as a function of time . Say x(t) =g(t)

    Now vavg = [x(t2)-x(t1)]/(t2-t1) =[x(12)-x(10)]/2

    Just give it a try.
     
  6. Oct 14, 2013 #5
    Thank you, but I still don't know if I think in a right way.
    So, I wrote dv= (150sin((2∏t)/30)-40)/8 dt
    ∫dv (from v0 - v)=∫((from 0-t) (150sin((2∏t)/30)-40)/8) dt
    If a is not constant do I have to integrate it, too?
    If yes, I get:
    v= ((-2250*cos((∏t)/15))/∏ -40t)/8
    If I place t=11s I get v=4,9 m/s

    Fot vavg I calculated v(t=10s) and v(t=12s) and got 3,6m/s. What is strange is, that v(t=10s)= - 5,24m/s.

    I really don't understand this concept well. -.-
     
  7. Oct 14, 2013 #6
    150Sin(2πt/30) - 40 = 8dv/dt

    [150Sin(2πt/30) - 40]dt = 8dv

    dv = (150/8)Sin(2πt/30)dt - (40/8)dt

    Integrate on both the sides .What do you get ?

    Note:Assuming the box starts from rest,the initial velocity is zero.
     
  8. Oct 16, 2013 #7
    Hi,

    I said that ∫(from 0 to 11) dv = ∫(from 0 to 11) (150/8)Sin(2πt/30)dt - (40/8)dt and I got (1125(1+sin7∏/30))/4∏ - 55 from that v(t=11s)=94,4m/s which is very strange to me. -.-
    I really can't find the solution. Can you please explain this exercise to me to the end, I must hand it in tomorrow. :eek:
     
  9. Oct 17, 2013 #8
    Hello, today I used wolfram alpha to integrate and I got:
    v(t)= (-Fdtω-F0cos(tω)+F0)/(ω*m)......v(11s)=94,43m/s
    After pulling: -Fd*t/m

    x(t)= (tω(2F0-Fdtω)-2F0sin(tω))/(2mω2)
    x(11s)=364,6m
    After pulling: -Fd*t2/(2m)
    x(40s)=-4km

    vavg=Δx/Δt.....463-275/2=94m/s

    I asked my friend for help and he got the same, but those results are very strange. ?!?!?! :eek:
     
  10. Oct 17, 2013 #9
    I simply cant follow you .You need to have more clarity while expressing your work . I am assuming you have written v(t=11s)=94.4m/s and not 94,4m/s . Why does it look strange to you ? The answer is what you get mathematically as per the question .

    What prompts you to calculate x(40s) ?

    Anyways, we have

    dv = (150/8)Sin(2∏t/30)dt - (40/8)dt

    ∫dv = ∫(150/8)Sin(2∏t/30)dt - ∫5dt

    v(t) = -(4500/16∏)Cos(2∏t/30) - 5t + C ,where C is constant of integration

    Now ,v(0) = 0 ,gives C = 4500/16∏

    So,v(t) = -(4500/16∏)Cos(2∏t/30) - 5t + 4500/16∏

    Now write v(t)=dx/dt ,

    dx/dt = -(4500/16∏)Cos(2∏t/30) - 5t + 4500/16∏

    dx = -(4500/16∏)Cos(2∏t/30) - 5t + 4500/16∏

    ∫dx = -∫(4500/16∏)Cos(2∏t/30)dt - ∫(5t)dt + ∫(4500/16∏)dt

    What value of x(t) do you get ? Please do not plug in the numbers .Just write down your expression of x(t) clearly .
     
    Last edited: Oct 17, 2013
  11. Oct 17, 2013 #10
    Like Tanya already mentioned, if that is the result that comes out after correctly solving it then why is it strange????......if you have specific doubt regarding the result, post it. BTW avg velocity does come out to be 94m/s.
     
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: Velocity from F(t) graph
Loading...