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GS-3 Section III Question 14

GS-3 Section III Question 14

Postby admin » Wed Aug 11, 2010 3:48 pm

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Re: GS-3 Section III Question 14

Postby Laura_2707 » Wed Mar 03, 2021 11:05 pm

Hi,

I'm just wondering for q14 how you get the answer to be (M1+M2+M3)g.

when filled into the F=MA equation I get: (M1+M2+M3)g = MA

So to get A on its own I must divide by M1 and then would end up with (M2+M3)g?

Thank you
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Re: GS-3 Section III Question 14

Postby goldstanda3269 » Sat Mar 06, 2021 5:10 am

Laura_2707 wrote:Hi,

I'm just wondering for q14 how you get the answer to be (M1+M2+M3)g.

when filled into the F=MA equation I get: (M1+M2+M3)g = MA

So to get A on its own I must divide by M1 and then would end up with (M2+M3)g?

Thank you


There are several issues with the reasoning above, first the maths:

If
(M1+M2+M3)g = MA
to isolate A, you must divide both sides by M:
(M1+M2+M3)/M x g = A
but if you meant M to be M1 then
(M1+M2+M3)/M1 x g = A (the M1's cannot cancel)

However, this statement is not correct:
(M1+M2+M3)g = MA
because: how is M1, which is on the ground, have its acceleration affected by gravity? It is not as direct as you have it above (it must involve friction, and so the coefficient of friction).

And the funny part: This question needs no calculation, the work was already done. Here is the key, from the passage: "The external force F is such that m3 neither rises nor falls." If m3 is not going up or down, then m2 is not moving relative to m1. That means that the acceleration of m2 is identical to that of m1. And since we already calculated the acceleration of m2 in the previous question, the acceleration of m1 is the same.

PS: You are making good progress through the practice material! I'm hoping that you have done, or are doing, the same progress through the GAMSAT-level practice questions in the books, while taking brief notes (and, studying from those notes 3 times/week, from the start, until your exam date). Keep up the good work! :)
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