A partially filled water tank is carried on a truck accelerating forward at a constant rate on level ground. What happens to the water free surface as seen in the tank?

Introduction / Context:Free-surface behavior in accelerating containers is a classic application of dynamic equilibrium in non-inertial frames. Understanding the tilt direction is important in vehicle-tank design, slosh analysis, and safe transport of liquids.

Given Data / Assumptions:

• Truck accelerates forward with constant horizontal acceleration a on level ground.
• Tank is partially filled; free surface is exposed to air.
• Quasi-steady condition, negligible wave dynamics during the instant considered.

Concept / Approach:In the accelerating frame, a pseudo body force acts on the liquid opposite to the direction of acceleration. The effective gravity vector g′ is the vector sum of real gravity g (downward) and pseudo acceleration a (backward). The free surface aligns perpendicular to g′, so it tilts upward toward the direction of the pseudo acceleration (toward the rear of the truck).

Step-by-Step Solution:Define g′ = g (down) + (−a) (backward).The free surface normal is along g′; therefore the surface slopes with tanθ = a/g.Since g′ leans backward, the liquid level is higher at the rear and lower at the front.

Verification / Alternative check:Small-angle prediction tanθ = a/g matches measurements in accelerating lab carts and explains spillage risk at the front baffle during rapid starts.

Why Other Options Are Wrong:Rise at front / fall at back: opposite to pseudo-force direction.Remains horizontal: only true when a = 0.None of these: incorrect because option stating rise at back, fall at front is correct.

Common Pitfalls:

• Confusing truck braking (deceleration) with acceleration; braking would reverse the tilt.
• Forgetting that the free surface is always perpendicular to the resultant of gravity and inertial forces.

Final Answer:It rises at the back side and falls at the front side