# Solve for one unknown mass in a simple pulley system [closed]

My class was given a problem where we would use a pulley system to calculate the mass of a pair of car keys. We attached 200g weights to each side of a pulley system and attached the keys to one of the weights. We would release the weight from a known height and measure the time for it to reach the ground to calculate acceleration. We also weighed the keys on a scale, so I know they actually weigh 37g

Using the acceleration, we are instructed to find the mass of the unknown weight. We are supposed to assume the pulley is massless and frictionless. These are my known variables:

$a=0.34m/s^2$

$m_1=0.2kg$

$m_2>m_1$

I made my free body diagrams, wrote my net force equations, and solved for T so I can use Newton's Second Law later on.

First:

$F_{net_{y1}}=T-F_{g_1}=m_1a$

$F_{net_{y1}}=T-m_1g=m_1a$

$T=m_1a+m_1g$

Second:

$F_{net_{y2}}=F_{g_2}-T=m_2a$

$T-m_2g=-m_2a$

Substitution:

$(m_1a + m_1g)-m_2g=-m_2a$

$m_1a+m_1g=m_2(-a + g)$

$$m_2=\frac{(m_1a + m_1g)}{(g - a)}$$

Plugging in values:

$$m_2=\frac{((0.2kg\times0.34m/s^2)+(0.2kg\times9.8m/s^2))}{(9.8m/s^2-0.34m/s^2)}$$

$m_2=0.214 kg$

Subtracting the 200g weight from that side, the calculated mass of the keys is 14g, when it is actually 37g. That is a 158% error...which is pretty bad. Is there a mistake in my equations, or did we collect some data that was just bad?

## closed as off-topic by David Z♦Jan 8 '15 at 6:16

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• I've done a lot more work, and I think my work is right here. Could the weight difference be insignificant enough that the energy loss due to the weight and the friction of the pulley is relatively large? – Seth Jan 8 '15 at 4:10
• How did you measure the acceleration? You can do a reverse calculation to find $a$. I doubt that the acceleration might be wrongly measured. – Jolie Jan 8 '15 at 4:19
• your equations are good , either $a$ is measured incorrectly or the pulleys/ropes have significant mass/friction. – Gowtham Jan 8 '15 at 4:22
• The acceleration could very well be inaccurate as well. We released the heavier side from .6 meters multiple times, and consistently got 1.89 seconds for the drop time. We then used $a = \frac{2dx}{t^2}$ to find acceleration. – Seth Jan 8 '15 at 4:24
• Thank you for verifying my equations. I will talk to my teacher tomorrow. I think my group may have used a set of keys that was too light to offset the energy loss due to the mass and friction of the rope and pulleys. – Seth Jan 8 '15 at 4:27