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When we consider a block placed on a table, we usually consider that the force of gravity acting downwards on the block (its weight) is balanced by the normal reaction from the table acting upwards on the block.

Why do we not consider a force due to air pressure to be exerted on the top surface of the block? Would the normal force then equal the net downward force on the block due to both its weight as well as the atmospheric pressure?

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Would the normal force then equal the net downward force on the block due to both its weight as well as the atmospheric pressure?

There's still air underneath the block, so air pressure acts on the block in all directions, cancelling out.

It's different for a suction cup. In that case a (partial) vacuum is created between the cup and a smooth surface and a net, downward force from atmospheric pressure acts on it. The total force would then be that force plus the weight of the cup.

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  • $\begingroup$ Good answer. As an addition to that, in the case of the suction cup, the increase in normal force that the cup exerts on the table is cancelled out by the decrease in total air pressure force on the top of the table (due to a vacuum in that area), so the system of table and cup of course has the same total weight. (Except for the small buoyant force due to archimedes principle). $\endgroup$
    – Azzinoth
    Jan 24 at 19:32

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