# more force problems

• Jun 28th 2006, 01:31 PM
Candy
The Atwood machine consists of two masses suspended from a fixed pulley. It is named after British scientist George Atwood (1746 - 1807), who used it to study motion and to measure the value of g. if m1 = 0.55kg and m2 = 0.80 kg, (a) what is the acceleration of the system, and (b) what is the magnitude of the tension in the string?
• Jun 28th 2006, 01:38 PM
Quick
For questions like this, that just require research, try looking on wikipedia.
• Jun 28th 2006, 04:29 PM
topsquark
Quote:

Originally Posted by Candy
The Atwood machine consists of two masses suspended from a fixed pulley. It is named after British scientist George Atwood (1746 - 1807), who used it to study motion and to measure the value of g. if m1 = 0.55kg and m2 = 0.80 kg, (a) what is the acceleration of the system, and (b) what is the magnitude of the tension in the string?

I can't draw the diagrams. Sorry!

This is an exercise in free body diagrams and keeping consistent coordinate directions.

I will assume m1 is hanging on the left side and m2 is on the right.

The FBD for m1:
I am choosing a +y direction upward. (This will be the direction of the acceleration of m1. I chose this because I know the answer. However you have to pick SOME direction to be positive. If you choose wrong your acceleration will simply be negative in the coordinate system, which still tells you what direction the acceleration is in.)
So we've got a tension in the string T pointing upward, and a weight w1 downward. Thus Newton's second says:
$\displaystyle \sum F_1 = T - w_1 = m_1 a$
We also know that w1 = (m1)g so this condition reads:
$\displaystyle T - m_1 g = m_1 a$

The FBD for m2:
Here's where you need to be careful. I am going to pick the +y direction to be downward. This is NOT because I know the answer, but because I defined the acceleration of m1 to be upward in the previous diagram. If m1 accelerates up m2 must accelerate down in order to be consistent.
Again we have a tension T pointing upward (which is the same magnitude as the tension in the previous diagram because we are using a massless string) and a weight w2 downward. The acceleration of m2 will be the same as the acceleration of m1 because the string is assumed not to stretch.
$\displaystyle \sum F_2 = w_2 - T = m_2 a$ (+y downward!)
Again, w2 = (m2)g so this condition reads:
$\displaystyle m_2 g - T = m_2 a$

We have two equations in two unknowns (a and T). Plugging the numbers in (and dropping units for convenience):
$\displaystyle T - 0.55g = 0.55a$
$\displaystyle 0.80g - T = 0.80a$

Solving the top equation for T I get:
$\displaystyle T = 0.55a + 0.55g$
and upon inserting this T value into the second equation:
$\displaystyle 0.80g - (0.55a + 0.55g) = 0.80a$

Solving this for a gives me:
$\displaystyle a = \frac{0.80g - 0.55g}{0.80 + 0.55} \approx 1.81 \, m/s^2$
(As a quick check, this acceleration must be less than g. We see that it is.)

Now to get T, use this value for a in either of the two original equations. I'll use the first equation:
$\displaystyle T - 0.55g = 0.55(1.81481)$ (I'm carrying a couple extra digits for rounding error purposes)
T = 6.39 N.

-Dan