# What force accelerates a liquid moving in a narrowing pipe?

You are right. From continuity of the incompressible fluid you have $$A_1 v_1 = A_2 v_2.$$ So obviously the velocity is changing. Thus the fluid is accelerated, and therefore there must be a force causing this acceleration. In this case the force comes from the pressure difference between the wide and the narrow part of the pipe.

(image from ResearchGate - Diagram of the Bernoulli principle)

This can be described by Bernoulli's equation ($$p$$ is pressure, $$\rho$$ is density) $$\frac{1}{2}\rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$

There is more mass per area behind than ahead of the constriction so since ppressure is force divided by area there develops a pressure difference

As you state, from the continuity equation you can see that the velocity $$v_2>v_1$$. The next step is a momentum balance (like any balance in fluid dynamics: $$\frac{d}{dt}=in-out+production$$). The momentum flowing into the system is smaller than the momentum flowing out of the system ($$\rho A_1 v_1^2 < \rho A_2 v_2^2 = \rho A_1 v_1^2 \frac{A_1}{A_2}$$, $$\frac{A_1}{A_2}>1$$).

The actual force is not the pressure itself, but the pressure difference, or, actually, the difference in force, because on the left the force is $$p_1 A_1$$ and on the right $$p_2 A_2$$.

From another conceptual point. Suppose you have a garden hose. Than you have a fixed pressure drop. Now, if you squeeze it, you create a contraction. Part of the pressure drop is now needed to accelerate the fluid at the exit. The overall flowrate decreases, because you also need pressure to overcome frictional forces.