Question

in the discussion question for lesson 12, makeshift scale, we considered creative ways to estimate mass from known parameters such as the masses of other objects and measurable distances. with that in mind, consider the figure of a mobile.
if the mass of the chicken is 23 grams and the donkey is 40 grams, estimate the distances ( x_2 ) and ( x_4 ) and the masses of the cow and pig, given that the total mass of the animals is slightly over 150 grams.

Explanation:

Step1: Calculate total mass of known animals

The known masses are chicken (\(m_{chicken} = 23\) g) and donkey (\(m_{donkey} = 40\) g). Let \(m_{cow}\) be mass of cow, \(m_{pig}\) be mass of pig. Total mass \(M = m_{chicken}+m_{donkey}+m_{cow}+m_{pig}>150\) g. So \(m_{cow}+m_{pig}>150 - 23 - 40=87\) g.

Step2: Use torque balance for the lowest level (donkey and chicken)

Torque balance: \(m_{donkey} \times X_1 = m_{chicken} \times X_2\). Given \(X_1 = 15\) cm, \(m_{donkey}=40\) g, \(m_{chicken}=23\) g. So \(40\times15 = 23\times X_2\). Solve for \(X_2\): \(X_2=\frac{40\times15}{23}=\frac{600}{23}\approx26.09\) cm.

Step3: Use torque balance for pig, donkey and chicken level

Let \(m_{pig}\) be mass of pig. The torque from pig side: \(m_{pig} \times X_3\), and from donkey - chicken side: \((m_{donkey}+m_{chicken}) \times X_4\). Given \(X_3 = 10\) cm, \(m_{donkey}+m_{chicken}=23 + 40 = 63\) g. So \(m_{pig}\times10=(23 + 40)\times X_4\). Also, from the next level (cow and pig - donkey - chicken), torque balance: \(m_{cow} \times X_5=(m_{pig}+m_{donkey}+m_{chicken}) \times X_6\). Given \(X_5 = 35\) cm, \(X_6 = 28\) cm. So \(m_{cow}\times35=(m_{pig}+63)\times28\). Simplify: \(m_{cow}=\frac{28}{35}(m_{pig}+63)=\frac{4}{5}(m_{pig}+63)\).

Step4: Combine with total mass constraint

Total mass: \(m_{cow}+m_{pig}+63>150\) \(\Rightarrow m_{cow}+m_{pig}>87\). Substitute \(m_{cow}=\frac{4}{5}(m_{pig}+63)\) into it: \(\frac{4}{5}(m_{pig}+63)+m_{pig}>87\). Multiply by 5: \(4(m_{pig}+63)+5m_{pig}>435\) \(\Rightarrow 4m_{pig}+252 + 5m_{pig}>435\) \(\Rightarrow 9m_{pig}>183\) \(\Rightarrow m_{pig}>\frac{183}{9}\approx20.33\) g. Also, from torque balance \(m_{pig}\times10=(63)\times X_4\), and from step 2, we can assume \(X_4\) is related. Let's assume \(m_{pig}\) such that total mass is slightly over 150. Let's try \(m_{pig}=30\) g (trial). Then from \(m_{pig}\times10 = 63\times X_4\), \(X_4=\frac{30\times10}{63}=\frac{300}{63}\approx4.76\) cm (not reasonable, wrong trial). Try \(m_{pig}=45\) g. Then \(X_4=\frac{45\times10}{63}=\frac{450}{63}\approx7.14\) cm. Then \(m_{cow}=\frac{4}{5}(45 + 63)=\frac{4}{5}\times108 = 86.4\) g. Total mass: \(23+40 + 45+86.4 = 194.4\) g (too big). Try \(m_{pig}=35\) g. \(X_4=\frac{35\times10}{63}=\frac{350}{63}\approx5.56\) cm. \(m_{cow}=\frac{4}{5}(35 + 63)=\frac{4}{5}\times98 = 78.4\) g. Total mass: \(23+40+35 + 78.4=176.4\) g (still big). Try \(m_{pig}=25\) g. \(X_4=\frac{25\times10}{63}\approx3.97\) cm. \(m_{cow}=\frac{4}{5}(25 + 63)=\frac{4}{5}\times88 = 70.4\) g. Total mass: \(23+40+25 + 70.4 = 158.4\) g (slightly over 150, good).

Step5: Verify torque for cow and pig - donkey - chicken level

\(m_{cow}\times35=70.4\times35 = 2464\). \((m_{pig}+63)\times28=(25 + 63)\times28=88\times28 = 2464\). So torque balance holds.

Step6: Calculate \(X_4\) with \(m_{pig}=25\) g

From \(m_{pig}\times10=(m_{donkey}+m_{chicken})\times X_4\), \(25\times10 = 63\times X_4\), so \(X_4=\frac{250}{63}\approx3.97\) cm (or using the earlier relation, but with \(m_{pig}=25\) g, it's consistent).

Answer:

• \(X_2\approx26.1\) cm, \(X_4\approx4.0\) cm (approximate), \(m_{cow}=70.4\) g, \(m_{pig}=25\) g (total mass \(158.4\) g, slightly over 150 g). (Note: Values may vary slightly based on approximation, but the method uses torque balance (lever principle) for mobile equilibrium.)