Hydraulic 35301

The hydraulic jack has pistons with a capacity of 6 cm² and 360 cm². How much force do we have to apply to the smaller piston to be able to lift a body weighing 180 kg on the larger piston?

Correct answer:

F₁ =  30 N

Step-by-step explanation:

$$\begin{gathered} S_1=6{\text{cm}}^2 \\ {S}_2=360{\text{cm}}^2 \\ {m}_2=180\text{kg} \\ g=10{\text{m/s}}^2 \\ {F}_2=g\cdot m_2=10\cdot 180=1800\text{N} \\ {p}_1=p_2 \\ {F}_1/S_1=F_2/S_2 \\ {F}_1=F_2\cdot \frac{S_1}{S_2}=1800\cdot \frac{6}{360}=\frac{1800\cdot 6}{360}=\frac{10800}{360}=30\text{N} \end{gathered}$$

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