Question

question 4
left piston (input)\tright piston (output)
volume\t\t
area\t5 sq.in.\t35 sq. in.
distance\t43 in\t
what is the output distance?

Explanation:

Step1: Recall Hydraulic Principle

In a hydraulic system, the volume input equals the volume output. Volume \( V = \text{Area} \times \text{Distance} \), so \( A_1 \times d_1 = A_2 \times d_2 \).

Step2: Substitute Known Values

Let \( A_1 = 5 \) sq.in, \( d_1 = 43 \) in, \( A_2 = 35 \) sq.in. We need to find \( d_2 \).
From \( A_1d_1 = A_2d_2 \), solve for \( d_2 \): \( d_2=\frac{A_1d_1}{A_2} \).
Substitute values: \( d_2 = \frac{5 \times 43}{35} \).

Step3: Calculate the Result

\( 5\times43 = 215 \), then \( \frac{215}{35}=\frac{43}{7}\approx6.14 \) (or keep as fraction \( \frac{43}{7} \)).

Answer:

The output distance is \( \frac{43}{7}\approx6.14 \) inches (or \( 6\frac{1}{7} \) inches).