2 Cubic Blocks on a Right Triangle

1. Jan 18, 2015

postfan

1. The problem statement, all variables and given/known data
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

2. Relevant equations
F_gravity =mg

3. 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?

2. Jan 18, 2015

Bystander

Good.
Good.
Good.
"Up?" Which "up?" Relative to what planes? There are three planes for the triangular block, and one for the table. Pick.

3. Jan 19, 2015

postfan

Up relative to the surface of the table.

4. Jan 19, 2015

Bystander

What components of the forces on the small blocks do you get when you multiply by cosα and cosβ?

5. Jan 19, 2015

postfan

The vertical components.

6. Jan 19, 2015

Bystander

What are the vertical forces on the small blocks?

7. Jan 19, 2015

postfan

Gravity acts with mg down and there is a normal force of m*g*cos(alpha or beta) up.

8. Jan 19, 2015

Bystander

"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. Jan 19, 2015

postfan

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. Jan 19, 2015

Bystander

Implies what about alpha and beta?
Reduces to what?

11. Jan 19, 2015

postfan

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. Jan 19, 2015

ehild

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. Jan 19, 2015

Bystander

Yes.

14. Jan 19, 2015

postfan

Ok so is the answer C?

15. Jan 19, 2015

Bystander

Yes.

16. Jan 19, 2015

ehild

Sure. :)

17. Jan 19, 2015

postfan

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. Jan 19, 2015

Bystander

Good conclusion.

19. Jan 19, 2015

postfan

Thanks!

20. Jan 19, 2015

BvU

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.