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Block with friction on acc' wedge

  1. Jul 4, 2011 #1
    A block rests on a wedge inclined at an angle theta. The coefficient of friction between the block and the plane is mu. (a) Find the maximum value of theta for the block to remain motionless on the wedge when the wedge is fixed in position. (b) The wedge is given horizontal acceleration a, as in the attachment. Assume that tan(theta) < mu, find the minimum acceleration for the block to remain on the wedge without sliding. Ans, clue if [tex] \theta = \frac{\pi}{4} \mbox{, } a_{min} = \frac{g(1-\mu}{1+\mu}[/tex] (c) repeat part b for the maximum value of acceleration.

    (a) At the point of slipping [tex] F_{friction} = w\sin\theta[/tex]. N the normal reaction of the plane decreases as [tex] N = w\cos\theta[/tex] The cos(theta) will decrease as theta increases. In the limit, when the block is at the point of slipping N = w*cos(theta), then
    [tex] \frac{F_{friction}}{N} = \frac{w\sin\theta}{w\cos\theta} = \tan\theta[/tex]. In the limit theta becomes the angle of friction [tex] \frac{F}{N} = \tan\theta = \mu [/tex]. Is part (a) answered ok? For parts b and c I am interested to know how to approach this problem. As you see [tex] a_{min}(1+\mu) = g(1-\mu) [/tex] implies that [tex] a_{min}\mu [/tex] is a valid component of the frictional force F_s.
    I am trying to envisage the acceleration a_min, I thought that a_min =0 as opposed to a_max which I think requires [tex] ma_{max}\cos\theta = F_s, or F_s\cos\theta = ma_{max}[/tex] from the free body diagram. Please clarify a_min as opposed to a_max?
    [tex] ma = (mg\sin\theta)\cos\theta + N\sin\theta - F_s\cos\theta[/tex]
    Where a is the acceleration of the block and wedge horoizontally. Is this equation correct?
    Don't I have to find how N is related to a and g, e.g., [tex] N=m\sqrt{g^2 - a^2}[/tex] Of course in this instance ma is taken parallel to the ramp. though! Please guide me with part (b). Thanks in advance.

    3. The attempt at a solution
     

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    Last edited: Jul 4, 2011
  2. jcsd
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