U-tube manometer in accelerating vehicle

Data: arms $b=60$ mm apart (fore-and-aft), alcohol density $\rho=0.79\times1000=790\ \text{kg/m}^3$, slope $\theta=20^\circ$ down, acceleration $a=2\ \text{m/s}^2$ forward (down the slope), level difference along the arms $=73$ mm, front limb higher.

Use axes fixed to the vehicle: $x$ along the slope (forward), $y$ perpendicular to the slope (along the arms). In this frame the fluid feels an effective body force per unit mass $\vec g-\vec a$:

• $x$ component: $g\sin\theta-a=9.81\sin20^\circ-2=1.355\ \text{m/s}^2$ (forward)
• $y$ component: $-g\cos\theta=-9.81\cos20^\circ=-9.218\ \text{m/s}^2$

Pressure varies as $dp=\rho(b_x\,dx+b_y\,dy)$ within the liquid. Go from the surface in the rear limb to the surface in the front limb: $\Delta x=+0.060$ m forward and $\Delta y=+0.073$ m (the front level is higher):

$$p_{front}-p_{rear}=790\left[1.355(0.060)-9.218(0.073)\right]=790(0.0813-0.6729)=-467.4\ \text{Pa}$$

The air pressure acting on each liquid surface is therefore lower in the front limb:

$$p_{rear}-p_{front}=467\ \text{Pa}$$

Answer: the air pressure difference is about 467 Pa (0.467 kPa), the rear limb being at the higher pressure.