Contact Force on ground problem fireman

I worked out part (i) using the volume of the cylinder and then multiplying it by the density of the water, which gave an exact answer of $8.25 kg s^{-1}$

In part (ii), this is my method thus far: The ground exerts a force to balance the fireman's weight (Newton's third law):

$$W = mg = 92 \times 9.81 = 902.52N$$

Now, I recognised there would be a vertical component of the force of water, which I thought was weight of the water, but then I realised that it wasn't acting on the fireman itself.

I then realised Newton's second law: $F = \frac{\triangle p}{\triangle t}$ where t = 1 so I then worked out the force needed to accelerate the water from rest per second:

$$F = m \triangle v = 8.25 \times 25 = 206.25N$$ The marking scheme accepts this, so the vertical force of the water is $206.25 sin55 = 168.95...$ N and then adds both forces together to give the answer:

$$168.95 + 902.52 = 1071.4 ... = 1100N$$ to 2 s.f.

But I'm confused. Surely the force needed to accelerate the water has a vertical component acting upwards, so it doesn't contribute to the contact force on the fireman's feet?

Can someone explain?

• Please consider to edit the question to focus on the conceptual part. In the moments the numbers and images of the question sheet hide the conceptual part of the question (why is the force for accelerating the water acting on the object holding the hose). As it currently stands people might confuse it for a "solve my homework" type of question, while this site will only answer conceptual questions about physics. – Sebastian Riese May 28 '18 at 14:33

It is a common misconception that the force on the nozzle is the same as the thrust on a rocket. Water is ejected through the nozzle but it is not ejected (thrown) by the nozzle. The water has already been accelerated (ejected/thrown) far back in the hose, at the pump. That is where the thrust occurs as the water is accelerated. There is no change in momentum of the water in the straight part of the nozzle so there is no force there. See Backward Reaction Force on a Fire Hose, Myth or Reality?

The downward and forward (not backward) force which the fireman feels is the force on hose caused by the change in direction of the water - which is also a change of momentum.

The water is delivered to the fireman horizontally. If he keeps the hose horizontal and the nozzle does not taper, then there is no force at all on the hose or nozzle as the water is ejected. In fact, if the nozzle tapers to increase the velocity of the ejected water, then the water exerts a forward force on the nozzle and hence on the fireman.

But in this problem the fireman re-directs the water upwards, bending the hose. The water exerts a reaction force on the hose equal and opposite to its rate of change of momentum. The fireman has to provide an upward and backward force to prevent the hose from unbending.
The above diagram shows the change in momentum of $8.25kg$ of water in the hose, which occurs ever second, from $p_1=206.25Ns$ horizontally to $p_2=206.25Ns$ at $55^{\circ}$ to the horizontal. The force on the water is in the direction of $\Delta p$, and the reaction force on the hose is in the opposite direction.

The vertical change in momentum per second $\Delta p_y$ is $169N$ as calculated using the thrust equation. But this is not because the water is leaving the hose at $55^{\circ}$ to the horizontal, it is because the water has been re-directed through an angle of $55^{\circ}$. If the initial momentum of the water $p_1$ had been say $25^{\circ}$ below the horizontal while the nozzle remained at $55^{\circ}$ above it, the vertical change in momentum would be greater.

Note that the horizontal change in momentum of the water $\Delta p_x$ is backwards (ie to the right) so the reaction force on the hose is forward (to the left). This is the opposite direction to that predicted by the incorrect use of the thrust equation.

Another force which has been omitted from the calculation in part (ii) is the weight of the hose and the water in it which the fireman is supporting. The weight of the water could be obtained by estimating the length of hose which is not supported by the ground (we already know the diameter, assuming this is the same as the nozzle). No information is given about the weight of the hose and nozzle.