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!

Gravitational Acceleration

  1. Dec 5, 2007 #1
    1. The problem statement, all variables and given/known data
    a block of mass (m) is placed on a bigger triangular shaped block with a mass (M). The wedge is of angle (α) . The coefficient of static friction between the two blocks is μs . What must the force (F) be in order to keep the smaller block at the same height on the wedge-shaped block? The answer is an equation in terms of the coefficient of friction, M, m, the angle and the gravitational acceleration. A FBD is also required. All laws must be documented and used properly.
    [​IMG]


    2. Relevant equations
    Newton's 1st Condition
    Gravitational Acceleration
    Definition of Friction


    3. The attempt at a solution

    I have the answer, but I'm having trouble getting to it.
    The answer to the problem:
    F = (M+m)((sinα-μscosα/cosα-μssinα))g
     
    Last edited: Dec 5, 2007
  2. jcsd
  3. Dec 5, 2007 #2
    You are missing another important theory, Newton's 2nd. Also, you should draw a free-body diagram to clearly communicate all of the forces being exerted on this body.

    Casey
     
  4. Dec 5, 2007 #3

    rl.bhat

    User Avatar
    Homework Helper

    The acceleration of the system = a = F/(M + m )
    the forces acting on the mass m are mg and ma. Resolve them into the components parallel and perpendicular to the surface of the wedge and find the condition for the equilibrium.
     
  5. Dec 6, 2007 #4
    Yeah that's where I'm stuck. I'm not sure where all of the forces apply on the two blocks, so my free body diagram is wrong. It's not the equations that are troubling me. It's getting all of the parts to the FBD that is troubling me.
     
  6. Dec 6, 2007 #5

    rl.bhat

    User Avatar
    Homework Helper

    I'm not sure where all of the forces apply on the two blocks Both the blocks experiance equal force in thr horizontal direction. You have to draw FBD for the body of mass m. As I have already mensioned, resolve them into the components parallel and perpendicular to the surface of the wedge and find the expression for a.
     
  7. Dec 6, 2007 #6
    mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)

    forces acting on m in equilibrium.
    forces due to gravity = forces due to F.
     
  8. Dec 6, 2007 #7

    rl.bhat

    User Avatar
    Homework Helper

    mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)
    That is correct. You can simplify this as g[Sin(alpha)-μsCos(alpha)]=a[Cos(alpha)-μsSin(alpha)]. Now substitute the value af a
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gravitational Acceleration
Loading...