2 Cubic Blocks on a Right Triangle

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Homework Statement


A right-triangular wooden block of mass M is at rest on a table, as shown in figure. Two smaller wooden cubes,
both with mass m, initially rest on the two sides of the larger block. As all contact surfaces are frictionless, the
smaller cubes start sliding down the larger block while the block remains at rest. What is the normal force from
the system to the table?
(A) 2mg
(B) 2mg + Mg
(C) mg + Mg
(D) Mg + mg(sin α + sin β)
(E) Mg + mg(cos α + cos β)

A picture can be found here : http://www.aapt.org/physicsteam/2014/upload/exam1-2013-1-6-unlocked.pdf
Page 4 Problem 11

Homework Equations


F_gravity =mg

The Attempt at a Solution


I did a FBD getting a force of mg down on the cubic blocks and Mg on the triangle block and a normal force of mgcos(alpha) and mgcos(beta) up respectively on the cubic blocks and Mg up on the triangle block. I added up all the forces to get (E) but the right answer is (C). Where did I go wrong?
 

Answers and Replies

  • #3
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Up relative to the surface of the table.
 
  • #4
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What components of the forces on the small blocks do you get when you multiply by cosα and cosβ?
 
  • #5
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The vertical components.
 
  • #6
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What are the vertical forces on the small blocks?
 
  • #7
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Gravity acts with mg down and there is a normal force of m*g*cos(alpha or beta) up.
 
  • #8
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"Normal" means perpendicular to the direction of motion of the small blocks. It does not mean "perpendicular to the table top" for the particular decomposition of vectors you're doing when you multiply by cosα and cosβ for this step in solving the problem.
 
  • #9
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Ok so won't decomposing the vector again to the axes defined as x being the surface of the table and y perpendicular to that make the normal force for the 2 blocks to = mgcos(alpha)sin(beta)+mgcos(beta)sin(alpha) respectively?
 
  • #10
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all contact surfaces are frictionless
larger block while the block remains at rest
Implies what about alpha and beta?
normal force for the 2 blocks to = mgcos(alpha)sin(beta)+mgcos(beta)sin(alpha)
Reduces to what?
 
  • #11
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Doesn't that reduce to just mg using the product to sum formulas so the total normal force is just Mg+mg. As a side note is there anyway I can do this without knowing the trig identities since I'm preparing for a contest that doesn't give you them.
 
  • #12
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Doesn't that reduce to just mg using the product to sum formulas so the total normal force is just Mg+mg. As a side note is there anyway I can do this without knowing the trig identities since I'm preparing for a contest that doesn't give you them.
You know that α+β=90°so sinα = cosβ and cosα=sinβ, so mg(cos(α)sin(β)+cos(β)sin(α))=mg (sin2(β)+cos2(β))
 
  • #13
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  • #14
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Ok so is the answer C?
 
  • #15
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Yes.
 
  • #17
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Cool thanks for the help! I think the lesson here is that all your forces have to be on the same axis for you to start comparing them even if you have to decompose them more than once.
 
  • #18
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Good conclusion.
 
  • #20
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As a side note is there anyway I can do this without knowing the trig identities since I'm preparing for a contest that doesn't give you them.
To some extent: yes. But it's a bit artificial, because you do have to decompose the mg on each of the two into a normal force and one along the sliding plane. The latter don't affect block M, the two normal forces do, so then you have to add the two normal force vectors. If at that point you see that the two decomposition diagrams are congruent with the one addition diagram, you can claim that the vector sum is mg without having to do the trig.

As I said, a bit artificial.
 

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