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I came across this example question in my mechanics book:

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It's a light scale pan with negligible mass, with two blocks A and B, with masses 0.4kg and 0.6kg respectively. The pan is attached to a string.

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A tension T (10.3N) is then applied to the string, causing the scale pan to accelerate at 0.5m/s/s.

The question is to work out the force exerted on B by A.

The book explains that the best way to go about this is to work out the force exerted on A by B, and then use Newton's 3rd Law to then say that A will exert an equal and opposite force on B.

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So it's just a simple matter of working out R, giving 4.12N, meaning 4.12N is exerted on A by B, thus meaning A exerts the same force on B.

What I don't understand this, is why do we ignore the gravitational acceleration of A? If A is sitting on B, does that not mean that A will also be exerting a force of 0.4g on B? Since this isn't the case, why isn't this so? And what would it entail if I did include A's gravitational acceleration?

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If the masses were not accelerating, it would be the case that B exerts an upward force on A equal to

$$F_{AB} = (0.4)\cdot(9.81) \mathrm N = 3.92 \mathrm N$$

in order to cancel the downward force of gravity.

Since A is accelerating upward, it must be that B exerts a greater force equal to

$$F_{BA} = (0.4) \cdot (9.81 + 0.5) = 4.12 \mathrm N $$

So gravity isn't ignored; B must exert a force equal to the weight of A plus the force required to accelerate the mass of A upward.

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  • $\begingroup$ Ok, but what about the 4.12N that A then exerts on b. What's the significance of that? Does that ensure that it remains seated on B? $\endgroup$
    – 83457
    Commented Jan 26, 2016 at 13:53

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