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Jan 12 An ATWOOD's Machine

M2 is shown to be bigger, but it isn't

Shown above is an ATWOOD's Machine. The question is: Given m1 = 6 KG and m2 = 4 KG, what is the acceleration of the system?

NOTE: We make an important assumption, which is that the pulley is massless.

MathJax TeX Test Page We look at each mass separately. We know $F_{net} = ma$, so we can plug in what we already know. The only forces are gravity and tension. We just have to figure out the direction of each block.

Mass 1

Mass 1 (6 kg) is more massive than Mass 2 (4 kg), so the net force is gravity subtracted by tension.

MathJax TeX Test Page $$F_g - T = ma$$ $$6(9.8) - T = 6a$$ $$T = 58.8 - 6a$$

Mass 2

Mass 2 loses the tug-of-war, because it's less massive. Mass 2 still has two forces, tension and gravity. Tension wins now, so the net force equals tension subtracted by gravity.

MathJax TeX Test Page $$T - F_g = ma$$ $$T - 4(9.8) = 4a$$ $$T = 39.2 + 4a$$

Now, we have two expressions for tension that we can set equal to each other to find the acceleration.

MathJax TeX Test Page $$T = 58.8 - 6a$$ $$T = 39.2 + 4a$$ $$58.8 - 6a = 39.2 + 4a \to 10a = 19.6$$ $$\boxed{a = 1.96 \text{ }m/s^2}$$

Now, which way is it going? This is very simple to check: just look where the mass is greater. Mass 1 is greater, so the system will accelerate to the left at 1.96 m/s^2.

The Pulley has Mass

For the problem above, we assumed the pulley has no mass. However, assume the pulley is extremely massive. That should surely slow down the acceleration, right? Let's figure out what the acceleration would be if the pulley had mass 12 kg.

MathJax TeX Test Page This now requires three equations. First, this involves torque. $$\tau_{net} = I\alpha$$ I is the moment of inertia. $\alpha$ is the angular acceleration. $\tau$ is the torque, or rotational force. $$F_{net} = ma$$ $$a = \alpha r$$ Now, let's start plugging things in. There are two different tensions within that string. The friction make it as if there is a third mass tied in the middle of the rope. The only reason the tension was the same through was because the pulley was completely massless. $$T_1r - T_2r = \dfrac{mr^2}{2}\dfrac{a}{r}$$ $$T_1 - T_2 = \dfrac{ma}{2}$$ Now, each point mass's acceleration is determined by tension and gravity. For mass 1, $$F_g - T_1 = ma$$ $$6(9.8) - T_1 = 6a$$ $$\boxed{T_1 = 58.8 - 6a}$$ Now, we do this for mass 2. $$T_2 - 4(9.8) = 4a \to \boxed{T_2 = 39.2 + 4a}$$ $$T_1 - T_2 = 19.6 - 10a = \dfrac{(12)a}{2}$$ $$19.6 - 10a = 6a \to a = \dfrac{19.6}{16} = \boxed{1.225 \text{ }m/s^2} \text{ to the left}$$

David Witten