Hydraulic lift

Martin needs to lift a 1.5-ton car using a hydraulic device with pistons of sizes 2 cm² and 15 dm². How much force must the device exert on the small piston?

Correct answer:

F₁ =  20 N

Step-by-step explanation:

$$\begin{gathered} S_1=2{\text{cm}}^2\to {\text{m}}^2=2:10000{\text{m}}^2=0.0002{\text{m}}^2 \\ {S}_2=15{\text{dm}}^2\to {\text{m}}^2=15:100{\text{m}}^2=0.15{\text{m}}^2 \\ {m}_2=1.5\text{t}\to \text{kg}=1.5\cdot 1000\text{kg}=1500\text{kg} \\ g=10{\text{m/s}}^2 \\ {F}_2=m_2\cdot g=1500\cdot 10=15000\text{N} \\ {p}_1=p_2 \\ {F}_1/S_1=F_2/S_2 \\ {F}_1\cdot S_2=F_2\cdot S_1 \\ {F}_1=F_2\cdot \frac{S_1}{S_2}=15000\cdot \frac{0.0002}{0.15}=20\text{N} \end{gathered}$$

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