Single-reverse 78404

Hydraulic jack has a capacity of 10 tons. The hydraulic lifter has 6 cm² and 360 cm² pistons. Determine the diameter of the small piston (d) and the force I will create on the piston (F). Design a single-reverse and a double-reverse lever.

Correct answer:

d =  2.764 cm
F₁ =  1666.6667 N

Step-by-step explanation:

$$\begin{gathered} S_1=6{\text{cm}}^2 \\ {S}_2=360{\text{cm}}^2 \\ {S}_1=\pi \cdot (d/2{)}^2 \\ d=2\cdot \sqrt{S_1/\pi }=2\cdot \sqrt{6/3.1416}=2.764\text{cm} \end{gathered}$$

$$\begin{gathered} m_2=10\text{t}\to \text{kg}=10\cdot 1000\text{kg}=10000\text{kg} \\ g=10{\text{m/s}}^2 \\ {F}_2=m_2\cdot g=10000\cdot 10=100000\text{N} \\ {p}_1=p_2 \\ {F}_1/S_1=F_2/S_2 \\ {F}_1=F_2\cdot \frac{S_1}{S_2}=100000\cdot \frac{6}{360}=\frac{100000\cdot 6}{360}=\frac{600000}{360}=\frac{5000}{3}\text{N}=1666.6667\text{N} \end{gathered}$$

Try another example