Physics Help Forum 2 blocks on top of each other exerting pressure on the ground

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 Sep 18th 2018, 10:21 AM #1 Junior Member   Join Date: Oct 2017 Posts: 8 2 blocks on top of each other exerting pressure on the ground 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?
 Sep 18th 2018, 11:15 AM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 703 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...) Latinized likes this. __________________ ~\o/~
 Sep 18th 2018, 01:38 PM #3 Senior Member   Join Date: Aug 2010 Posts: 384 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 likes this.
 Sep 18th 2018, 04:35 PM #4 Junior Member   Join Date: Oct 2017 Posts: 8 Ah ok, the bottom block was M1 and A1 so its b). But thank you guys so much!

 Tags blocks, exerting, ground, pressure, top

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