Question

Here is a standard tire gauge, shown as a cross-section (from ) There are two environments, separated by a movable piston. To the right of the piston is a spring and rod, both open to the room atmospheric pressure. To the left of the piston is open to the tire chamber, so will be at the tire's pressure when connected. Assume that when the gauge is not connected, the gauge reads zero. (This is called a differential pressure gauge, and is the typical gauge). Here are the specifications of the gauge: Piston area = 7.9 times 10^-5 m^2. Spring constant of the spring = 9.24 times 10^2 N/m. Assume your garage is at 1atm. NOW you connect the gauge to your motorcycle and the piston forces the rod out a distance of 2.0cm (0.020 meters). Obtain the differential pressure in the tire, in psi. (The differential pressure is the amount of pressure the tire is above 1atm.) Obtain the actual pressure in the tire. That is, include the atmospheric pressure.

Answer 1

spring constant k = 9.24 x 10² N/m

when the rod moves out through a distance 2.0 cm, compression in the spring x is 2.0 cm.

Tension in the spring is T = kx = 18.48 N

Area of piston A = 7.9 x 10^-5 m²

(a) differential pressure P_d = T/A = 2.34 x 10⁵ N/m² = 2.34 x 10⁵ Pa = 33.9 psi

(b) actual pressure = differential pressure + atmospheric pressure = 33.9 + 14.7 = 48.6 psi