Calculating Error: Expression for Mass of Pulley m_P

[temaire]

Homework Statement

The expression for the slope of the graph of $$(m_{1} - m_{2})$$ versus a is given by $$m = (m_{1} + m_{2} + m_{p})/g$$. What is the expression for the error in the mass of the pulley, $$m_{P}$$?

The Attempt at a Solution

My answer is $$\delta_{p} = g\delta_{m} + \delta_{m1} + \delta_{m2}$$

I don't think this is right. Can someone show me where I went wrong?

 

[Mark44]

Homework Statement

The expression for the slope of the graph of $$(m_{1} - m_{2})$$ versus a is given by $$m = (m_{1} + m_{2} + m_{p})/g$$. What is the expression for the error in the mass of the pulley, $$m_{P}$$?

The Attempt at a Solution

My answer is $$\delta_{p} = g\delta_{m} + \delta_{m1} + \delta_{m2}$$

I don't think this is right. Can someone show me where I went wrong?

None of this makes any sense, as far as I can tell. In your first sentence you say

The expression for the slope of the graph of $$(m_{1} - m_{2})$$ versus a is given by $$m = (m_{1} + m_{2} + m_{p})/g$$.

If you are graphing m₁ - m₂ vs. a, why doesn't a appear in the equation?

 

[temaire]

I'm not sure if this would help, but $$(m_{1}-m_{2})g = (m_{1}+m_{2})a$$

I think g might have been substituted for a.

 

[Mark44]

Here's where I think this is going. You have m = (1/g)(m₁ + m₂ + m_p), so
$$dm = \frac{\partial d m}{\partial m_1}\Delta m_1 + \frac{\partial d m}{\partial m_1}\Delta m_1 +\frac{\partial d m}{\partial m_1}\Delta m_1$$
$$= (1/g)[1 \Delta m_1 + 1 \Delta m_2 + \Delta m_p]$$

Now solve for $\Delta m_p$ in terms of the other quantities.