A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7 m/s^2
The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest.
(a) Assuming rigid-body acceleration of the coffee, determine whether it will spill out of the mug. (b) Calculate the gage pressure in the corner at point A if the density of coffee is 1010 kg/m^3
• System sketch: Figure E2.13 shows the coffee tilted during the acceleration.
• Assumptions: Rigid-body horizontal acceleration, a_{x} = 7 m/s^2. Symmetric coffee cup.
• Property values: Density of coffee given as 1010 kg/m^3
• Approach (a): Determine the angle of tilt from the known acceleration, then find the height rise.
• Solution steps: From Eq. (2.39), the angle of tilt is given by
\theta =\tan ^{-1}\frac{a_{x}}{g} ==\tan ^{-1}\frac{7.0 m/s^2}{9.81 m/s^2} =35.5°
If the mug is symmetric, the tilted surface will pass through the center point of the rest position, as shown in Fig. E2.13. Then the rear side of the coffee free surface will rise an amount \Delta z given by \Delta z=(3 cm)(\tan 35.5°)=2.14 cm \lt 3cm therefore no spilling
• Comment (a): This solution neglects sloshing, which might occur if the start-up is uneven.
• Approach (b): The pressure at A can be computed from Eq. (2.40)
, using the perpendicular distance \Delta s from the surface to A. When at rest, P_{A} = \rho gh_{rest} =(1010 kg/m^3) (9.81 m/s^2)(0.07 m) =694 Pa. When accelerating,
P_{A} = \rho G \Delta s =\left(1010\frac{kg}{m^3} \right) \left[\sqrt{(9.81)^2+(7.0)^2} \right] [3(0.07+ 0.0214)\cos 35.5°] \approx 906 Pa
Comment (b): The acceleration has increased the pressure at A by 31 percent. Think about this alternative: why does it work? Since a_{z}= 0, we may proceed vertically down the left side to compute
P_{A} = \rho g(z_{surf}-z_{A})=(1010 kg/m^3)(9.81 m/s^2)(0.0214+ 0.07 m)= 906 Pa
