# How is Normal Force on Table affected by two Blocks sliding on Triangle?

Suppose we have a triangle with mass M as shown:

Clearly, the normal force of from the system to the table is $Mg$.

But what if we added two blocks of equal mass m on sides BC and AC of triangle ABC? What would the normal force from the system (=triangle plus two blocks) to the table be then? (assuming all contact surfaces are frictionless, so the two blocks immediately start sliding off). Assume the triangle does not move on the surface it is on.

I believe the answer is $(2m+M)g$ but I am not sure if the two masses of mass $m$ that are sliding down the triangle affect the normal force in any way.

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The normal force would be $g(M + m(\cos^2 \theta_1 + \cos^2\theta_2))$
where the $\theta$'s represent the angles at the bottom of the triangles.
I got the precise numbers because I memorized that a block sliding frictionlessly down a slope has acceleration $g\sin\theta$. The vertical component of their accelerations is $g\sin^2\theta$. This is the amount that the normal force is reduced, so each block contributes an amount $mg - mg\sin^2\theta = mg \cos^2\theta$ to the normal force.
 I'll have to disagree with that. The correct expression for the normal force should be $(M+m(\cos^2 \theta_1 + \cos^2 \theta_2))g$. To see that your expression is incorrect, put $\theta_1 = \theta_2 = 0^{\circ}$: it gives $Mg$ instead of $(M+2m)g$. – Alraxite Jan 19 at 0:17 @Alraxite Your expression is nonsense, but mine was also wrong, so I'll update it. – Mark Eichenlaub Jan 19 at 0:33 Your new expression is incorrect as well. The one I've given is correct. Would you mind explaining how did you arrive at that? – Alraxite Jan 19 at 0:43 Yeah, I made a mistake because I didn't stop to work it out carefully. Sorry to have been so dismissive, and thank you for pointing it out. – Mark Eichenlaub Jan 19 at 0:51