Question

A hydraulic crane supports a mass of 500 kg with its piston-cylinder actuator, inclined at 60° with horizontal, as shown in the figure below. The diameter of the piston is 30 cm and gravitational acceleration is 10 m/s\(^2\). Neglecting mass of the piston, the reading of the pressure gage in kPa (gage) is

• A. 61.26
• B. 6.13
• C. 35.35
• D. 70.71

Hint:

Resolve the weight of the load along the direction of the inclined piston rod before dividing by the piston area.

Answer 1

The Correct Option is A

Step 1: Set up the force balance along the piston axis.
A piston-cylinder can only push or pull along its own axis, so only the component of the weight along that axis is balanced by the oil pressure.
The weight acts straight down: \( W = mg = 500 \times 10 = 5000 \) N.
The axis is inclined at \( 60^{\circ} \) to the horizontal, so it makes \( 30^{\circ} \) with the vertical, and the component of W along the axis is \( W \sin 60^{\circ} \).

Step 2: Compute the axial force and the piston area.
Axial force \( F = 5000 \times \sin 60^{\circ} = 5000 \times 0.8660 = 4330.1 \) N.
Piston diameter \( D = 0.30 \) m, so area \( A = \dfrac{\pi}{4} D^2 = \dfrac{\pi}{4}(0.30)^2 = 0.07069 \ \text{m}^2 \).

Step 3: Find the gage pressure.
\( P = \dfrac{F}{A} = \dfrac{4330.1}{0.07069} = 61257 \) Pa \( \approx 61.26 \) kPa.

Final Answer:
The gage pressure needed to hold the 500 kg mass on the inclined piston is about 61.26 kPa. \[ \boxed{P \approx 61.26 \text{ kPa}} \]