Mass — Question 3

Question 3

Reversing a balance reveals an unequal-arm error

A balance has fixed positive arm lengths $L$ and $R$, which may differ. For this model, after pan contributions are removed, a level balance obeys $$(\text{left mass})L=(\text{right mass})R.$$ The local gravitational field is the same at both pans. Ignore buoyancy, friction, and all other errors.

An unknown object balances $S_1=121.00\ \mathrm{g}$ of standards when the object is on the left. With the object on the right, it balances $S_2=100.00\ \mathrm{g}$ on the left. Treat these as exact model data.

1. Write the two balance equations and eliminate $L/R$ to find the unknown mass.

2. Infer $L/R$. Which arm is longer, and why does the first comparison need more standard mass than the object actually has?

3. A technician averages 121.00 g and 100.00 g. Find that estimate’s error and explain why reversal does not justify this arithmetic average.

4. Name one instrumental effect excluded by the model that reversal alone would not necessarily remove.

Question 3 - Solution

Reversing a balance reveals an unequal-arm error

(a) Eliminate the fixed arm ratio.

The first configuration gives $mL=S_1R$; the second gives $S_2L=mR$. Dividing the first equation by the second cancels the nonzero arm lengths and gives $m/S_2=S_1/m$. Hence $$m^2=S_1S_2,\qquad \boxed{m=\sqrt{(121.00)(100.00)}\ \mathrm{g}=110.00\ \mathrm{g}}.$$ Only the positive root represents a mass.

(b) Recover the arm ratio and its direction.

$$\frac LR=\frac{S_1}{m}=\sqrt{\frac{S_1}{S_2}}=\sqrt{\frac{121}{100}}=1.10.$$

From the first equation, $\boxed{L/R=S_1/m=121/110=1.10}$. The left arm is longer. The object on that arm produces the same balance effect as a larger standard mass on the shorter right arm.

(c) The arm error is multiplicative, not equal and opposite.

$$\frac{S_1+S_2}{2}-\sqrt{S_1S_2}=\frac{(\sqrt{S_1}-\sqrt{S_2})^2}{2}=0.50\ \mathrm g.$$

The arithmetic estimate is $(121+100)/2=110.50\ \mathrm{g}$, an overestimate of $\boxed{0.50\ \mathrm{g}}$ (approximately 0.455%). The two comparisons scale the unknown by reciprocal factors: $S_1=mL/R$ and $S_2=mR/L$. Their product is $m^2$, but their sum is not generally $2m$.

(d) Reversal corrects only the effects modeled.

Examples include a changing zero torque, position-dependent friction, or unequal buoyancy forces. Such effects add terms or change the balance relation; the two equations used above would need modification. Merely exchanging pans does not guarantee their cancellation.