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By what percentage is a true weight of the girls being marked up by the shop keeper?

By what percentage is a true weight of the girls being marked up by the shop keeper-example-1

1 Answer

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ANSWER

3. 5.93432

Step-by-step explanation

The pans are 67.6cm apart. If the fulcrum was centered, the distance from each pan to the fulcrum would be,


\frac{67.6\operatorname{cm}}{2}=33.8\operatorname{cm}

Now, the fulcrum was moved 0.974cm towards one of the pans,

The new distances are,


\begin{gathered} d_1=33.8cm+0.974\operatorname{cm}=34.774cm \\ d_2=33.8cm-0.974\operatorname{cm}=32.826cm \end{gathered}

When the shopkeeper weighs the products, the balance is in equilibrium. This means that the torque each weight produces must be equal,


\begin{gathered} \tau_1=\tau_2 \\ F_1d_1=F_2d_2_{} \end{gathered}

The relationship between the forces is,


F_2=(d_1)/(d_2)F_1

Replace with the distances,


\begin{gathered} F_2=\frac{34.774\operatorname{cm}}{32.826\operatorname{cm}}\cdot F_1 \\ \\ F_2=1.0593432F_1 \end{gathered}

Now we have to find the percent change,


(F_2-F_1)/(F_1)\cdot100\%=(1.0563432F_1-F_1)/(F_1)\cdot100\%=0.0593432\cdot100\%=5.93432\%

Hence, the true weight of the goods is being marked up by 5.93432%.

By what percentage is a true weight of the girls being marked up by the shop keeper-example-1
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