Inclined plane and friction

  • #1

Homework Statement



There are two wedges, each with mass m, placed on a flat floor. A cube of mass M is balanced on the wedges. Assume there is no friction between wedge and block, and the coefficient of static friction is less than 1 between the floor and wedges. What is the largest M that can be balanced without moving?

diagram here:
http://sites.google.com/site/learningphys/

Homework Equations


The Attempt at a Solution



If we split the block in half due to symmetry, the question will get easier.

force of friction is given by:
[tex]F=\mu (\frac{M}{2}+m)g[/tex]
the M/2 comes from taking only half of the block

i want to equate the force that the block has on one of the wedge.

Still considering half of the diagram, i want the normal force of (M/2)g to the wedge's surface. The normal force is give by the usual inclined plane equation, mgcos(theta)

So normal force of (M/2) is:

[tex]N=\frac{Mg}{2}cos(\theta)[/tex]

this normal force pushes in the opposite direction of the force of friction... but we have to consider the horizontal component, which tags on a sin(theta) term

I finally get:

[tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2}cos(\theta)sin(\theta)[/tex]

And then i solve for mu. How close am I to being correct?
 

Answers and Replies

  • #2
tiny-tim
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Assume there is no friction between wedge and block,

Still considering half of the diagram, i want the normal force of (M/2)g to the wedge's surface. The normal force is give by the usual inclined plane equation, mgcos(theta)

So normal force of (M/2) is:

[tex]N=\frac{Mg}{2}cos(\theta)[/tex]
Hi learning_phys! :smile:

No … consider the forces on the block

there is no friction, so the only forces are Mg/2 and N …

the horizontal components balance anyway …

so resolve vertically, and that gives you … ? :smile:
 
  • #3
are you sayin N=Mg/2? but the normal force should be perpendicular to the blocks surface right? (that's why I have the cosine term)
 
  • #4
tiny-tim
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but the normal force should be perpendicular to the blocks surface right?
Yes (because there is no friction). :smile:
are you sayin N=Mg/2?
No … resolve vertically.
 
  • #6
tiny-tim
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  • #7
awesome

so I would get:
[tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2cos(\theta)}sin(\theta )[/tex]
 
  • #8
tiny-tim
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awesome

so I would get:
[tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2cos(\theta)}sin(\theta )[/tex]
:biggrin: Woohoo! :biggrin:
(you're easily awed! … :wink:)
And so the largest M is … ? :smile:
 

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