Latinized

So the question is that there are 2 blocks on top of each other of different mass (M1 and M2) and different cross sectional areas (A1 and A2). They are lying on a smooth surface. What would be the pressure exerted by the blocks on the surface.

a) (M1g/A1) +(M2g/A2)

b) (M1g/A1) +(M2g/A1)

c) (M1g/A2) + (M2g/A2)

d) (M1g/A1) - (M2g/A2)

There is no answer given. However, I'm bit confused why there is no option of ((M1+M2)*g)/(A1+A2). Since Fg = Fn and P = Fn/A, the normal force of the blocks would be (M1+M2)*g. That would be divided by the sum of the areas. How would I solve this?

Woody

Which is the top block?
Which is the bottom block?

The idea of combining the 2 masses in your equation is sensible,
but note that only the bottom block is in contact with the surface.

(extra hint, your equation would be correct if the blocks were side by side...)

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HallsofIvy

The fact that there are two blocks and the base area of the top block are really irrelevant. There is a total weight of M1g+ M2g contacting the ground with area A2 (I am assuming that is the area of the lower block). The pressure is (M1g+ M2g)/A2= M1g/A2+ M2g/A2, your answer (c).

Latinized

Ah ok, the bottom block was M1 and A1 so its b). But thank you guys so much!