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A 5.00-kg object placed on a frictionless, horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging 9.00-kg object, as in Figure. Find the acceleration of the two objects and the tension in the string.

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The given answer looks like:

$$ mg-T=9a \\ 9 \times 9.8 - T = 9a \\ 22.2 - T = 9a \text{ --- (1)} $$

Where did 9a come from?

Then it continues

$$ T=5a \text{ --- (2)} \\ \text{sub (2) into (1)} \\ 22.2 - 5a = 9a \\ 14a = 88.2 \\ a = 6.3 \\ T = 5a = 5 \times 6.3 = 31.5N $$

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2 Answers 2

up vote 1 down vote accepted

Hint: Distinguish between the part of the system that is responsible for the gravitational force and the total system that is accelerated.

You can also suppose that the mass of the pulley and any friction phenomenon that could be associated to it, is neglected. Further suppose that the cable is inextensible.

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Hmm, am I right to say $mg - T$ is the downward net force on the 9KG block? Err... then $F=ma$ so $F = 9a$ and they are equals? Ah maybe I got it? –  Jiew Meng Sep 27 '12 at 11:58
    
Effectively, the net force on $m_1 = 9kg$ is given by $m_1g - T$, thus the 2nd law of Newton for $m_1$ writes: $m_1g - T = m_1a$. Now what remains to do, is to write Newton's 2nd law for the other mass $m_2 = 5kg$ to find out what is $T$ –  dedoco Sep 27 '12 at 13:32
    
By the way, I strongly advice you to not replace the masses and accelerations by the actual values, but to keep the symbols (so do not replace $m_1$ with 9). This really helps to not confuse things, also it allows you to check whether the units are right. –  dedoco Sep 27 '12 at 13:34

Let us assume m(2) accelerates downward. Force Body Diagram for m(1)

From the figure of m(1) we get, F(x)= T-m(1)a(x)=m(1)a F(y)=n-m(1)g=0=m(1)a(y)

Force body diagram for m(2)

From figure of m(2) we get,

F(y)=m(2)g-T=m(2)a(y)=m(2)a

solving the above equations we get,

m(2)g-m(1)a=m(2)a implies a={m(2)/(m(1)+m(2))}g implies a={9/5+9}*9.81=6.30m/s^2 T=m(1)a=31.5 N

I hope now it is clear from where 9a comes from. here,F(x),a(x) are the force and acceleration along x-direction and so on.

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