3 weights

A two-arm lever has a 0.5 kg weight on the left arm at a distance of 64 cm from the axis of rotation, and a 0.8 kg weight at a distance of 30 cm from the axis of rotation. How far from the axis of rotation on the right arm must a 0.7 kg weight be placed for the lever to be in equilibrium?

Final Answer:

r =  0.8 m

Step-by-step explanation:

$$\begin{gathered} m_1=0.5\text{kg} \\ {l}_1=64\text{cm}\to \text{m}=64:100\text{m}=0.64\text{m} \\ {m}_2=0.8\text{kg} \\ {l}_2=30\text{cm}\to \text{m}=30:100\text{m}=0.3\text{m} \\ {m}_3=0.7\text{kg} \\ g=9.81{\text{m/s}}^2 \\ {M}_1+M_2=M_3 \\ {F}_1\cdot l_1+F_2\cdot l_2=F_3\cdot r \\ {m}_1\cdot g\cdot l_1+m_2\cdot g\cdot l_2=m_3\cdot g\cdot r \\ {m}_1\cdot l_1+m_2\cdot l_2=m_3\cdot r \\ r=\frac{m_1\cdot l_1+m_2\cdot l_2}{m_3}=\frac{0.5\cdot 0.64+0.8\cdot 0.3}{0.7}=0.8\text{m} \end{gathered}$$

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