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!

Block slipping against wedge

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider a fixed wedge, whose part is a quarter circle of radius as shown. A small block of mass 'm' slides down the wedge and leaves horizontally from a height 'h' as shown.The coefficient of friction between wedge and block is 'μ[itex]_{k}[/itex]' . Compute the value of in centimetres , where 'x' is the distance from bottom of wedge of the point where the block falls given the values below:
    Here is the figure.
    https://d18l82el6cdm1i.cloudfront.net/solvable/9f0d7815b5.e8a0649017.E9Nd6g.png
    _Details : _

    R=200cm
    h=105cm
    μ[itex]_{k}[/itex]=0.5
    m=2g


    2. Relevant equations



    3. The attempt at a solution

    I think we have to apply law of conservation of energy here.
    i.e., potential energy of block
    (E[itex]_{p}[/itex]) =work done by friction (W[itex]_{f}[/itex]) + Kinetic energy (E[itex]_{k}[/itex])....(1)
    Here, I have problem in finding the W[itex]_{f}[/itex].
    In know F[itex]_{f}[/itex] = μR.
    Here, the normal reactionary force which is being exerted on the block on the wedge is not constant. I tried few ways but I am unable to find W[itex]_{f}[/itex]. It would be nice if somebody could help.
     
    Last edited: Aug 17, 2014
  2. jcsd
  3. Aug 17, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    And it depends on the velocity. If the acceleration depends on the velocity, a good idea is to set up a differential equation and try to solve that.
     
  4. Aug 17, 2014 #3
    Thanks for the quick reply mfb.

    I tried this--

    dw=μrdθ(mgsinθ+mrω[itex]^{2}[/itex])

    I know that ω=dθ/dt.

    But putting this does not help.
    Is there any other way through which we can find ω?
    pppppp8.png This is the figure.
     
    Last edited: Aug 17, 2014
  5. Aug 17, 2014 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    What is w?

    Can you express everything in terms of θ, ω and dω/dt?
     
  6. Aug 17, 2014 #5
    dw infinitely small work done by frictional force.
     
  7. Aug 17, 2014 #6
    I think I can.

    As a=αr and α=dω/dt
    gcosθ-gsinθ-rω^2=rdω/dt

    Is it right?
     
  8. Aug 17, 2014 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Work is not relevant here.

    Somehow the μ got lost, apart from that it looks good.
     
  9. Aug 17, 2014 #8
    gcosθ-μgsinθ-μrω^2=rdω/dt.

    What to do next?
     
  10. Aug 17, 2014 #9
    Is eq.1 in my post#1 correct?
     
  11. Aug 17, 2014 #10

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Do some mathematical magic to try to find a solution.
    As a homework problem, it should have a reasonable solution, I just don't see it now.
     
  12. Aug 17, 2014 #11

    CAF123

    User Avatar
    Gold Member

    Hi Satvik,
    Yes, it is. I think you forgot the contribution of the work done by gravity along the circular arc though in #3.

    I seem to be having the same problem as you. I can write down an explicit expression for the work done by all the external forces but there is one term in it that I cannot integrate.
     
  13. Aug 17, 2014 #12
    Hi CAF123
    Work done by gravity is mgrsinθ.Is it right?
     
  14. Aug 17, 2014 #13

    CAF123

    User Avatar
    Gold Member

    I think I should write out the problem I am having.

    The work done on the mass is due to friction and gravity. An infinitesimal displacement of the mass along the arc is ##\text{d} \underline s = R\theta \hat s## using the definition of ##\theta## in your attachment. The arc is now parametrised in terms of ##\theta## and so $$W = \int_0^{\pi/2} \underline{F} \cdot \text{d} \underline s = \underbrace{-\int_0^{\pi/2} \mu\left(mg \sin \theta + \frac{mv(\theta)^2}{R}\right)R\text{d}\theta}_{\text{work done by friction}}+ \underbrace{\int_0^{\pi/2} mg \cos \theta R \text{d}\theta}_{\text{work done by gravity}}$$ This can be rewritten like $$W = -R\mu mg \int_0^{\pi/2} \sin \theta \text{d} \theta - mR^2 \int_0^{\pi/2} \dot{\theta}^2 \text{d}\theta + mgR\int_0^{\pi/2} \cos \theta \text{d}\theta$$

    It is this middle integral that I cannot evaluate. The last integral is mgR, which we could have obtained simply by noting gravity is a conservative force and the mass falls a vertical distance R. The 'work done by friction' brace is the finite version of your dw I believe.
     
  15. Aug 17, 2014 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Try the trick of using the chain rule to write dω/dt = (dω/dθ)(dθ/dt)
     
  16. Aug 17, 2014 #15

    CAF123

    User Avatar
    Gold Member

    Hi TSny,
    Do you have any suggestions on how to compute the middle integral in my last post? Should I find ##\dot{\theta}## first?
     
  17. Aug 17, 2014 #16

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I don't see a way to do the integral. It seems to me that solving the differential equation in Satvik's post #8 is the way to go.

    If you like work and energy, you can get an equivalent differential equation by considering the differential form of the work-energy theorem: dW = d(mv2/2) and using your expressions for the work done by friction and gravity for an infinitesimal displacement ds = Rdθ. This way, you avoid getting a differential equation with a time derivative. (In Satvik's equation, you have a time derivative which can be rewritten in terms of a derivative with respect to θ using the chain rule.)

    θ is the independent variable of the diff. eq.
    For the dependent variable, y say, choose either y = v2 or y = ω2.
     
    Last edited: Aug 17, 2014
  18. Aug 17, 2014 #17

    CAF123

    User Avatar
    Gold Member

    Yes, so the equation becomes $$\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$$ So, I could solve this for ##\dot{\theta}(\theta)## and input into my W-E energy equation.
     
    Last edited: Aug 17, 2014
  19. Aug 17, 2014 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. But, of course, if you can solve the differential equation ##\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta## for ##\omega(\theta)##, you essentially have the answer for the speed at the bottom of the track. There is no need to go back to the W-E equation.
     
  20. Aug 17, 2014 #19

    CAF123

    User Avatar
    Gold Member

    Yes, of course, thanks. Could you elaborate on what you meant by the below argument?
    I could take the differential of the equation W, and I think then the first and third terms of my equation would vanish (they are constants) but I am not sure what the middle term would look like. Or did you mean something else?
    Thanks!
     
  21. Aug 17, 2014 #20

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Fwiw, maybe not much, I simulated it. Assuming the exit speed is proportional to sqrt(rg), plotting for mu up to 0.55 gives an excellent fit to ##v = \sqrt{rg} 1.9187(0.75 - \mu)##. No idea what the magic constant relates to. Presumably the 0.75 indicates the threshold at which it sticks somewhere on the wedge.
    Goes a bit wrong for higher mu, but that might be rounding errors.
     
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: Block slipping against wedge
  1. Block on Wedge (Replies: 25)

  2. WEDGE and block (Replies: 3)

Loading...