Question

question 13 (1 point)
a 13,000-n vehicle is to be lifted by a 25-cm diameter hydraulic piston. what force needs to be applied to a 5.0 cm diameter piston to accomplish this? assume the pistons each have negligible weight.

a) 260 n

b) 2600 n

c) 520 n

d) 5200 n

Explanation:

Step1: Recall Pascal's Principle

Pascal's Principle states that pressure is transmitted equally in a fluid, so \( P_1 = P_2 \), where \( P = \frac{F}{A} \) (pressure equals force over area). Thus, \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \).

Step2: Calculate Areas of Pistons

The area of a circle is \( A = \pi r^2 = \pi (\frac{d}{2})^2 \). For the large piston (d₂ = 25 cm), \( A_2 = \pi (\frac{25}{2})^2 \). For the small piston (d₁ = 5.0 cm), \( A_1 = \pi (\frac{5.0}{2})^2 \).

Step3: Relate Forces and Areas

From \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \), solve for \( F_1 \): \( F_1 = F_2 \times \frac{A_1}{A_2} \). Substitute \( A_1 \) and \( A_2 \): \( \frac{A_1}{A_2} = \frac{(\frac{5.0}{2})^2}{(\frac{25}{2})^2} = (\frac{5.0}{25})^2 = (\frac{1}{5})^2 = \frac{1}{25} \).

Step4: Compute \( F_1 \)

Given \( F_2 = 13,000 \, \text{N} \), \( F_1 = 13,000 \times \frac{1}{25} = 520 \, \text{N} \)? Wait, no—wait, wait, correction: Wait, \( A_1 \) is small, \( A_2 \) is large. Wait, \( F_2 \) is the force on the large piston (lifting the vehicle), so \( F_1 \) is the force on the small piston. Wait, \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \) → \( F_1 = F_2 \times \frac{A_1}{A_2} \). \( A_1 = \pi (2.5)^2 \), \( A_2 = \pi (12.5)^2 \). So \( \frac{A_1}{A_2} = (\frac{2.5}{12.5})^2 = (\frac{1}{5})^2 = \frac{1}{25} \). Then \( F_1 = 13,000 \times \frac{1}{25} = 520 \, \text{N} \)? Wait, but let's check again. Wait, diameter 25 cm: radius 12.5 cm; diameter 5 cm: radius 2.5 cm. Area ratio is (2.5/12.5)² = (1/5)² = 1/25. So force on small piston is 13,000 * (1/25) = 520 N? Wait, but the options have 520 N as option c. Wait, but let me recalculate: 13,000 divided by 25: 13,000 ÷ 25 = 520. Yes. Wait, but wait, maybe I mixed up F1 and F2. Wait, F2 is the force on the large piston (13,000 N), so F1 is the force on the small piston. So \( F_1 = F_2 \times \frac{A_1}{A_2} \). Since \( A_1 < A_2 \), F1 should be less than F2. 520 N is less than 13,000 N. So that makes sense.

Answer:

c) 520 N