Question

A cart of mass 10 kg is placed on a $60^\circ$ inclined plane. A turning vane (weight negligible) is mounted on the cart. A horizontal water jet from a 0.1 m$^2$ nozzle strikes the vane and is turned downward parallel to the inclined plane. Find the minimum jet velocity (m/s) so that the cart does not slide down. Neglect friction. Water density = 1000 kg/m$^3$, $g = 10$ m/s$^2$. (Round off to two decimal places)

Hint:

When a jet strikes a vane and changes direction, thrust equals mass flow rate times velocity change. Resolve forces parallel to the incline.

Answer 1

Correct Answer: 0.74

Weight component pulling cart down: \[ W \sin 60^\circ = 10 \times 10 \times \frac{\sqrt{3}}{2} = 86.60~\text{N} \] Momentum force from jet: \[ F = \rho A V^2 \] Here: $\rho = 1000$, $A = 0.1$ m$^2$. Thus: \[ F = 100 V^2 \] But force acts horizontally, its component opposing downslope motion is: \[ F_{\parallel} = F \cos 60^\circ = 100 V^2 \times 0.5 = 50 V^2 \] Equilibrium condition: \[ 50 V^2 = 86.60 \] \[ V^2 = 1.732 \] \[ V = 0.760~\text{m/s} \] \[ \boxed{0.76~\text{m/s}} \]