0
$\begingroup$

I came across this example question in my mechanics book:

enter image description here

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.

enter image description here

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.

enter image description here

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 Jan 26 '16 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.