System of particles

The explosion forces are internal, so the total linear momentum is conserved. Weights are proportional to masses, so the factor $1/g$ cancels and the weights can be used directly.

              A (5 lb)
               /  45 deg
 v0 =100 --->  *----------> x
               \  30 deg
              B (15 lb)

Before: the projectile moves horizontally at $100$ ft/s.

x-direction:

$$20(100)=5v_A\cos45^\circ+15v_B\cos30^\circ$$ $$2000=3.5355v_A+12.990v_B\qquad(1)$$

y-direction (initial vertical momentum is zero):

$$0=5v_A\sin45^\circ-15v_B\sin30^\circ$$ $$3.5355v_A=7.5v_B\ \Rightarrow\ v_A=2.1213v_B\qquad(2)$$

Substituting (2) in (1):

$$2000=3.5355(2.1213v_B)+12.990v_B=20.490v_B$$ $$v_B=97.61\ \text{ft/s},\qquad v_A=2.1213(97.61)=207.06\ \text{ft/s}$$

Answer: $v_A=207.1$ ft/s at $45^\circ$ above the horizontal; $v_B=97.6$ ft/s at $30^\circ$ below the horizontal.

Variable system of particles

A variable system of particles is one whose total mass changes with time because particles continuously join it or leave it, such as a rocket ejecting gas, a conveyor belt loaded with sand, or a chain being lifted from a heap. Its equation of motion is $\sum\vec F=m\vec a-\dfrac{dm}{dt}\vec u_{rel}$.

Exploding projectile

Internal forces of the explosion do not change the total momentum. Masses: $m=20$ kg, $m_A=5$ kg, $m_B=15$ kg, $v_0=30$ m/s (the numbers given in the question text; the figure shows the same arrangement in lb and ft/s).

              A (5 kg)
               /  45 deg
 v0 = 30 --->  *----------> x
               \  30 deg
              B (15 kg)

x-direction:

$$20(30)=5v_A\cos45^\circ+15v_B\cos30^\circ$$ $$600=3.5355v_A+12.990v_B\qquad(1)$$

y-direction:

$$0=5v_A\sin45^\circ-15v_B\sin30^\circ\ \Rightarrow\ v_A=2.1213v_B\qquad(2)$$

Substituting (2) in (1):

$$600=7.5v_B+12.990v_B=20.490v_B\ \Rightarrow\ v_B=29.28\ \text{m/s}$$ $$v_A=2.1213(29.28)=62.12\ \text{m/s}$$

Answer: $v_A=62.12$ m/s at $45^\circ$ above the horizontal; $v_B=29.28$ m/s at $30^\circ$ below the horizontal.

Angular momentum of a system of particles

The angular momentum (moment of momentum) of a system of $n$ particles about a fixed point O is the vector sum of the moments of the momenta of the particles:

$$\vec H_O=\sum_{i=1}^{n}\vec r_i\times m_i\vec v_i$$

About the mass centre G, using velocities relative to G or absolute ones, $\vec H_G=\sum\vec r'_i\times m_i\vec v'_i$. The two are related by $\vec H_O=\vec{\bar r}\times m\vec{\bar v}+\vec H_G$. Its rate of change equals the moment of the external forces: $\sum\vec M_O=\dot{\vec H}_O$. Unit: kg m$^2$/s.

Exploding projectile

Data: $m=10$ kg, $v_0=30$ m/s, $m_A=2.5$ kg, $m_B=7.5$ kg (the masses given in the question; the "1.5 kg" in the figure label is treated as a misprint because $2.5+7.5=10$). $\theta_A=45^\circ$ above, $\theta_B=30^\circ$ below the horizontal. Linear momentum is conserved.

x-direction:

$$10(30)=2.5v_A\cos45^\circ+7.5v_B\cos30^\circ$$ $$300=1.7678v_A+6.4952v_B\qquad(1)$$

y-direction:

$$0=2.5v_A\sin45^\circ-7.5v_B\sin30^\circ\ \Rightarrow\ v_A=2.1213v_B\qquad(2)$$

Using (2) in (1):

$$300=3.75v_B+6.4952v_B=10.245v_B\ \Rightarrow\ v_B=29.28\ \text{m/s},\quad v_A=62.12\ \text{m/s}$$

Answer: $v_A=62.1$ m/s at $45^\circ$ above the horizontal; $v_B=29.3$ m/s at $30^\circ$ below the horizontal.

Take the origin at the fixed pivot $O$, $x$ vertically downward, $y$ to the right and $\hat k$ out of the plane ($+\hat k$ = counter-clockwise). Assume both arms lie on the same (right-hand) side of the vertical, and that $\omega_2$ is the absolute angular velocity of the lower arm (as is usual in this problem). Angular momentum of the system about $O$:

$$\vec H_O = m_1(\vec r_1 \times \vec v_1) + m_2(\vec r_2 \times \vec v_2)$$

Position vectors

Upper arm ($a=0.5$ m at $30^\circ$ to the $x$-axis), lower arm ($b=0.7$ m at $60^\circ$ to the vertical):

$$\begin{aligned} \vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.433\,\hat i + 0.250\,\hat j)\ \text{m}\\ \vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.350\,\hat i + 0.606\,\hat j)\ \text{m}\\ \vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.783\,\hat i + 0.856\,\hat j)\ \text{m} \end{aligned}$$

Velocities

$$\begin{aligned} \vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.500\,\hat i + 0.866\,\hat j)\ \text{m/s}\\ \vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.319\,\hat i + 1.916\,\hat j)\ \text{m/s} \end{aligned}$$

Angular momentum

$$\begin{aligned} H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 1(0.433\times 0.866 - 0.250\times -0.500) = 0.50\ \text{kg m}^2/\text{s}\\ H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 1(0.783\times 1.916 - 0.856\times -2.319) = 3.49\ \text{kg m}^2/\text{s} \end{aligned}$$ $$\vec H_O = H_1 + H_2 = 3.99\,\hat k\ \text{kg m}^2/\text{s}$$

Answer: $\vec H_O \approx 3.99\,\hat k$ kg m²/s (counter-clockwise, about the axis perpendicular to the plane).

Take the origin at the fixed pivot $O$, $x$ vertically downward, $y$ to the right and $\hat k$ out of the plane ($+\hat k$ = counter-clockwise). Assume both arms lie on the same (right-hand) side of the vertical, and that $\omega_2$ is the absolute angular velocity of the lower arm (as is usual in this problem). Angular momentum of the system about $O$:

$$\vec H_O = m_1(\vec r_1 \times \vec v_1) + m_2(\vec r_2 \times \vec v_2)$$

Position vectors

Upper arm ($a=0.4$ m at $30^\circ$ to the $x$-axis), lower arm ($b=0.5$ m at $60^\circ$ to the vertical):

$$\begin{aligned} \vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.346\,\hat i + 0.200\,\hat j)\ \text{m}\\ \vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.250\,\hat i + 0.433\,\hat j)\ \text{m}\\ \vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.596\,\hat i + 0.633\,\hat j)\ \text{m} \end{aligned}$$

Velocities

$$\begin{aligned} \vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.800\,\hat i + 1.386\,\hat j)\ \text{m/s}\\ \vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.965\,\hat i + 2.636\,\hat j)\ \text{m/s} \end{aligned}$$

Angular momentum

$$\begin{aligned} H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 3(0.346\times 1.386 - 0.200\times -0.800) = 1.92\ \text{kg m}^2/\text{s}\\ H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 4(0.596\times 2.636 - 0.633\times -2.965) = 13.80\ \text{kg m}^2/\text{s} \end{aligned}$$ $$\vec H_O = H_1 + H_2 = 15.72\,\hat k\ \text{kg m}^2/\text{s}$$

Answer: $\vec H_O \approx 15.72\,\hat k$ kg m²/s (counter-clockwise, about the axis perpendicular to the plane).

Take the origin at the fixed pivot $O$, $x$ vertically downward, $y$ to the right and $\hat k$ out of the plane ($+\hat k$ = counter-clockwise). Assume both arms lie on the same (right-hand) side of the vertical, and that $\omega_2$ is the absolute angular velocity of the lower arm (as is usual in this problem). Angular momentum of the system about $O$:

$$\vec H_O = m_1(\vec r_1 \times \vec v_1) + m_2(\vec r_2 \times \vec v_2)$$

Position vectors

Upper arm ($a=0.5$ m at $30^\circ$ to the $x$-axis), lower arm ($b=0.6$ m at $60^\circ$ to the vertical):

$$\begin{aligned} \vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.433\,\hat i + 0.250\,\hat j)\ \text{m}\\ \vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.300\,\hat i + 0.520\,\hat j)\ \text{m}\\ \vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.733\,\hat i + 0.770\,\hat j)\ \text{m} \end{aligned}$$

Velocities

$$\begin{aligned} \vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.500\,\hat i + 0.866\,\hat j)\ \text{m/s}\\ \vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.578\,\hat i + 2.066\,\hat j)\ \text{m/s} \end{aligned}$$

Angular momentum

$$\begin{aligned} H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 1(0.433\times 0.866 - 0.250\times -0.500) = 0.50\ \text{kg m}^2/\text{s}\\ H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 2(0.733\times 2.066 - 0.770\times -2.578) = 7.00\ \text{kg m}^2/\text{s} \end{aligned}$$ $$\vec H_O = H_1 + H_2 = 7.50\,\hat k\ \text{kg m}^2/\text{s}$$

Answer: $\vec H_O \approx 7.50\,\hat k$ kg m²/s (counter-clockwise, about the axis perpendicular to the plane).

A variable-mass system gains or loses mass continuously. Consider a system of mass $m$ moving with velocity $\vec v$ that absorbs particles of mass $\Delta m$ moving with absolute velocity $\vec u$ during time $\Delta t$.

 before:  m ---> v        dm ---> u
 after:   (m + dm) ---> v + dv

Impulse-momentum for the whole set of particles

Momentum at $t$: $m\vec v+\Delta m\,\vec u$. Momentum at $t+\Delta t$: $(m+\Delta m)(\vec v+\Delta\vec v)$. External resultant force $\sum\vec F$ acts for $\Delta t$:

$$m\vec v+\Delta m\,\vec u+\sum\vec F\,\Delta t=(m+\Delta m)(\vec v+\Delta\vec v)$$

Neglecting the second-order term $\Delta m\,\Delta\vec v$ and dividing by $\Delta t$:

$$\sum\vec F=m\frac{d\vec v}{dt}+\frac{dm}{dt}(\vec v-\vec u)$$

Using the velocity of the absorbed particles relative to the system, $\vec u_{rel}=\vec u-\vec v$:

$$\boxed{\sum\vec F=m\vec a-\frac{dm}{dt}\,\vec u_{rel}}$$

or $m\vec a=\sum\vec F+\dfrac{dm}{dt}\vec u_{rel}$.

• Mass gained, particles initially at rest ($\vec u=0$): $\sum\vec F=\dfrac{d(m\vec v)}{dt}$.
• Mass ejected (rocket): $dm/dt<0$ and $\vec u_{rel}$ is the exhaust velocity relative to the body (backward), so the term $\dfrac{dm}{dt}\vec u_{rel}$ is a forward thrust $T=\left|\dfrac{dm}{dt}\right|u_{rel}$. Then $m\vec a=\sum\vec F+\vec T$.

Force due to change in mass

Let a system of mass $m$ moving with velocity $\vec v$ lose (eject) a small mass $\Delta m$ in time $\Delta t$, the ejected particles leaving with absolute velocity $\vec u$. External force $\sum\vec F$ acts.

Momentum at $t$: $m\vec v$. At $t+\Delta t$: $(m-\Delta m)(\vec v+\Delta\vec v)+\Delta m\,\vec u$. Impulse-momentum:

$$m\vec v+\sum\vec F\,\Delta t=(m-\Delta m)(\vec v+\Delta\vec v)+\Delta m\,\vec u$$

Neglecting $\Delta m\,\Delta\vec v$ and dividing by $\Delta t$ (with $\dot m=-\Delta m/\Delta t$ the rate of change of system mass, so $\dot m<0$):

$$\sum\vec F=m\frac{d\vec v}{dt}+\frac{\Delta m}{\Delta t}(\vec u-\vec v)$$

With $\vec u_{rel}=\vec u-\vec v$ (exhaust velocity relative to the body) and $\left|\dot m\right|=\Delta m/\Delta t$:

$$m\vec a=\sum\vec F-\left|\dot m\right|\vec u_{rel}$$

Here $\vec u_{rel}$ points backward, so $\vec T=-\left|\dot m\right|\vec u_{rel}$ is a forward thrust force. Hence $m\vec a=\sum\vec F+\vec T$ with $T=\left|\dot m\right|u_{rel}$.

Why the acceleration increases when mass is lost

Along the line of motion:

$$a=\frac{\sum F+T}{m(t)},\qquad m(t)=m_0-\left|\dot m\right|t$$

Take a constant burn rate and constant exhaust speed, so $T$ is constant and, for example, the external force (weight, drag) is roughly constant. The numerator stays the same while the denominator $m(t)$ decreases, so $a$ increases with time:

$$\frac{da}{dt}=\frac{(\sum F+T)\left|\dot m\right|}{m^{2}}>0$$

(when $\sum F+T>0$). The final acceleration is therefore larger than the initial one. For a rocket in free space ($\sum F=0$): $dv=u_{rel}\dfrac{-dm}{m}$, giving $v=v_0+u_{rel}\ln\dfrac{m_0}{m}$, so the velocity grows faster as $m$ becomes small.

Kinetic energy of a system of particles

Let the particle $i$ of mass $m_i$ have velocity $\vec v_i=\vec{\bar v}+\vec v'_i$, where $\vec{\bar v}$ is the velocity of the mass centre G and $\vec v'_i$ is the velocity relative to G. Then

$$T=\tfrac12\sum m_iv_i^{2}=\tfrac12\sum m_i(\vec{\bar v}+\vec v'_i)\cdot(\vec{\bar v}+\vec v'_i)$$ $$T=\tfrac12\bar v^{2}\sum m_i+\vec{\bar v}\cdot\sum m_i\vec v'_i+\tfrac12\sum m_iv_i'^{2}$$

The middle term is zero because $\sum m_i\vec v'_i=0$ (momentum relative to G). So

$$\boxed{T=\tfrac12m\bar v^{2}+\tfrac12\sum m_iv_i'^{2}}$$

The kinetic energy of the system equals the kinetic energy of the total mass moving with the mass centre plus the kinetic energy due to motion relative to the mass centre.

Water jet on a moving plate

Assumption: the plate is flat and perpendicular to the jet, and the water leaves along the plate (so all the relative velocity normal to the plate is lost). The plate moves at $v$ in the jet direction, so the mass of water that strikes it per second is based on the relative velocity $v_1-v$:

$$\dot m=\rho A(v_1-v)$$

The force on the plate equals the rate of change of the momentum of water:

$$P=\dot m(v_1-v)=\rho A(v_1-v)^{2}$$ $$400=1000(600\times10^{-6})(30-v)^{2}\ \Rightarrow\ (30-v)^{2}=666.7$$ $$30-v=25.82\ \Rightarrow\ v=4.18\ \text{m/s}$$

Answer: the plate speed is $v\approx4.18$ m/s.

Positions (figure only partly legible): the following position vectors, in metres, are assumed from the dimensions marked on the figure: $\vec r_A=1\hat i+3\hat j+4\hat k$, $\vec r_B=3\hat i+2\hat j+1\hat k$, $\vec r_C=1.5\hat i+1\hat j+1\hat k$. Replace them by the figure values if they differ; the method is unchanged. Masses 1, 2, 3 kg and the given velocities.

Momenta $m_i\vec v_i$ (kg m/s)

$$\vec p_A=3\hat i-2\hat j+4\hat k,\quad \vec p_B=8\hat i+6\hat j,\quad \vec p_C=6\hat i+15\hat j-9\hat k$$

(i) Angular momentum about O

$$\vec H_O=\sum\vec r_i\times m_i\vec v_i$$ $$\begin{aligned} \vec r_A\times\vec p_A&=(3(4)-4(-2))\hat i-(1(4)-4(3))\hat j+(1(-2)-3(3))\hat k=20\hat i+8\hat j-11\hat k\\ \vec r_B\times\vec p_B&=(2(0)-1(6))\hat i-(3(0)-1(8))\hat j+(3(6)-2(8))\hat k=-6\hat i+8\hat j+2\hat k\\ \vec r_C\times\vec p_C&=(1(-9)-1(15))\hat i-(1.5(-9)-1(6))\hat j+(1.5(15)-1(6))\hat k=-24\hat i+19.5\hat j+16.5\hat k \end{aligned}$$ $$\vec H_O=-10\hat i+35.5\hat j+7.5\hat k\ \ \text{kg m}^2/\text{s}$$

(ii) Mass centre G

$$\vec{\bar r}=\frac{\sum m_i\vec r_i}{\sum m_i}=\frac{1(1,3,4)+2(3,2,1)+3(1.5,1,1)}{6}=\frac{(11.5,\ 10,\ 9)}{6}$$ $$\vec{\bar r}=1.917\hat i+1.667\hat j+1.500\hat k\ \text{m}$$

(iii) Angular momentum about G

Mean velocity: $\vec{\bar v}=\dfrac{\sum m_i\vec v_i}{6}=\dfrac{(17,\ 19,\ -5)}{6}=2.833\hat i+3.167\hat j-0.833\hat k$ m/s. Total momentum $m\vec{\bar v}=17\hat i+19\hat j-5\hat k$.

$$\vec H_G=\vec H_O-\vec{\bar r}\times m\vec{\bar v}$$ $$\vec{\bar r}\times m\vec{\bar v}=-36.83\hat i+35.08\hat j+8.08\hat k\quad\Rightarrow\quad\vec H_G=26.83\hat i+0.417\hat j-0.583\hat k\ \ \text{kg m}^2/\text{s}$$

(Checked by computing $\sum(\vec r_i-\vec{\bar r})\times m_i(\vec v_i-\vec{\bar v})$ directly, giving the same result.)

Answer: $\vec H_O=(-10,\ 35.5,\ 7.5)$; $\vec{\bar r}=(1.917,\ 1.667,\ 1.5)$ m; $\vec H_G=(26.83,\ 0.417,\ -0.583)$ kg m$^2$/s (for the assumed positions).

Linear and angular momentum of a system of particles

• Linear momentum: $\vec L=\sum m_i\vec v_i=m\vec{\bar v}$, where $m$ is the total mass and $\vec{\bar v}$ the velocity of the mass centre. Newton's law: $\sum\vec F=\dot{\vec L}$.
• Angular momentum about O: $\vec H_O=\sum\vec r_i\times m_i\vec v_i$, with $\sum\vec M_O=\dot{\vec H}_O$.

Force of a fixed vane on a steady stream

Assumption: the stream arrives horizontally (to the right) and leaves at C at $60^\circ$ above the horizontal, in the forward direction. Since the vane is fixed and friction is neglected, the speed is unchanged: $v_B=v_C=60$ m/s.

   V_B --->   ______
              \      \  V_C (60 deg above horizontal)
               \______\
                 vane

Mass flow rate:

$$\dot m=\rho Av=1000(100\times10^{-6})(60)=6.0\ \text{kg/s}$$

Momentum equation (force on the stream by the vane):

$$\vec F=\dot m(\vec v_C-\vec v_B)$$ $$\begin{aligned} F_x&=6\left(60\cos60^\circ-60\right)=-180\ \text{N}\\ F_y&=6\left(60\sin60^\circ-0\right)=311.8\ \text{N}\\ F&=\sqrt{180^{2}+311.8^{2}}=360\ \text{N} \end{aligned}$$

The direction is $\tan^{-1}(311.8/180)=60^\circ$ above the negative x-axis (that is $120^\circ$ from +x). Check: $F=2\dot mv\sin(\tfrac{\alpha}{2})=2(6)(60)\sin30^\circ=360$ N, where $\alpha=60^\circ$ is the deflection angle.

Answer: the vane exerts $\vec F=(-180\hat i+311.8\hat j)$ N on the stream, magnitude $360$ N at $120^\circ$ to the original flow (the stream pushes on the vane with an equal and opposite force).

Assumptions: the vane moves at $V=16$ m/s in the direction of the jet; the stream leaves at C at $60^\circ$ above the horizontal relative to the vane; friction is neglected, so the relative speed is unchanged along the vane.

Relative velocity and flow rate

$$v_{rel}=v-V=48-16=32\ \text{m/s}$$

The mass of water striking a single moving vane per second is

$$\dot m=\rho A(v-V)=1000(4000\times10^{-6})(32)=128\ \text{kg/s}$$

Force on the stream

Apply momentum in the frame moving with the vane (constant velocity, so Newton's laws hold):

$$\vec F=\dot m(\vec v_{C,rel}-\vec v_{B,rel})$$ $$\begin{aligned} F_x&=128\left(32\cos60^\circ-32\right)=-2048\ \text{N}\\ F_y&=128\left(32\sin60^\circ\right)=3547\ \text{N}\\ F&=\sqrt{2048^{2}+3547^{2}}=4096\ \text{N} \end{aligned}$$

This is the force the vane exerts on the stream. The stream exerts an equal and opposite force on the vane: $2048$ N forward (along the jet) and $3547$ N downward.

Power developed

Only the force component in the direction of the vane's motion does work:

$$P=F_xV=2048(16)=32\,768\ \text{W}\approx32.8\ \text{kW}$$

Note that $V=v/3=16$ m/s is exactly the vane speed that makes the power of a single moving vane maximum (see the derivation $V=v/3$ for $P_{max}$), so this is the maximum power.

Answer: resultant force on the stream $=4096$ N (components $-2048$ N and $3547$ N); maximum power $\approx32.8$ kW.

Assumptions (figure not shown): the blade turns the water so that, relative to the blade, it leaves at $60^\circ$ above the horizontal with unchanged relative speed (no friction). The blade moves away from the jet at $5$ ft/s. Pound-force units.

Data

$$\rho=\frac{62.4}{32.2}=1.938\ \text{slug/ft}^3,\qquad A=\frac{\pi}{4}\left(\frac{2}{12}\right)^{2}=0.02182\ \text{ft}^2$$ $$v_{rel}=25-5=20\ \text{ft/s}$$

Mass flow rate hitting the moving blade

$$\dot m=\rho A\,v_{rel}=1.938(0.02182)(20)=0.8456\ \text{slug/s}$$

Force of the blade on the water

In the frame moving with the blade, the water enters at $20$ ft/s horizontally and leaves at $20$ ft/s at $60^\circ$:

$$\begin{aligned} F_x&=\dot m(20\cos60^\circ-20)=0.8456(-10)=-8.46\ \text{lb}\\ F_y&=\dot m(20\sin60^\circ)=0.8456(17.32)=14.65\ \text{lb} \end{aligned}$$

The blade pushes on the water with $8.46$ lb opposite to the jet direction (horizontal component) and $14.65$ lb upward (vertical component); resultant $16.9$ lb.

Power on the blade

The water pushes the blade forward with $8.46$ lb at speed $5$ ft/s:

$$P=8.456(5)=42.3\ \text{ft lb/s}=\frac{42.3}{550}=0.0769\ \text{hp}$$

Answer: $F_x=-8.46$ lb, $F_y=+14.65$ lb (on the water); power developed $\approx42.3$ ft lb/s ($0.077$ hp).

Setup. The jet has area $A$ and speed $V_A$. The single blade moves at $V$ in the jet direction. The water moves along the blade with constant speed relative to it, so its relative speed is $V_A-V$ at entry and at exit. It leaves at angle $\theta$ to the horizontal (relative to the blade).

  V_A --->  \
             \__  V_rel leaves at theta
   blade ->  V

i) Components of the force exerted by the blade on the stream

Mass of water reaching the moving blade per second: $\dot m=\rho A(V_A-V)$.

Use the frame of the blade ($V$ constant), with relative speed $V_A-V$:

$$F_x=\dot m\left[(V_A-V)\cos\theta-(V_A-V)\right]=-\rho A(V_A-V)^{2}(1-\cos\theta)$$ $$F_y=\dot m(V_A-V)\sin\theta=\rho A(V_A-V)^{2}\sin\theta$$

The negative $F_x$ means the force on the stream is opposite to the jet direction. The stream pushes the blade with $\rho A(V_A-V)^{2}(1-\cos\theta)$ forward and $\rho A(V_A-V)^{2}\sin\theta$ downward.

ii) Velocity for maximum power

Power developed on the blade (only the forward force does work):

$$P=F_{blade,x}\,V=\rho A(V_A-V)^{2}(1-\cos\theta)\,V$$

For maximum power, $dP/dV=0$:

$$\frac{d}{dV}\left[(V_A-V)^{2}V\right]=(V_A-V)^{2}-2V(V_A-V)=(V_A-V)(V_A-3V)=0$$ $$\boxed{V=\frac{V_A}{3}}$$

(the root $V=V_A$ gives zero power). Then

$$P_{max}=\rho A\left(\tfrac23V_A\right)^{2}(1-\cos\theta)\frac{V_A}{3}=\frac{4}{27}\rho AV_A^{3}(1-\cos\theta)$$

Answer: (i) $F_x=-\rho A(V_A-V)^2(1-\cos\theta)$, $F_y=\rho A(V_A-V)^2\sin\theta$ on the stream; (ii) power is maximum at $V=V_A/3$.

A steady stream is a continuous flow in which the same mass of particles passes through any section per second, so the momentum of the particles inside the control region (between the entry section A and the exit section B) does not change with time.

     entry A          vane           exit B
  v_A -->  ======== curved ========  --> v_B
           (mass flow m_dot in = out)

Derivation

Consider the particles in the portion of the stream between A and B at time $t$, plus the mass $\Delta m$ that enters at A during $\Delta t$.

• Momentum at $t$: $\vec H_{AB}+\Delta m\,\vec v_A$, where $\vec H_{AB}$ is the momentum of the particles between A and B.
• Momentum at $t+\Delta t$: $\vec H_{AB}+\Delta m\,\vec v_B$ (the mass $\Delta m$ has left at B; the contents between A and B are unchanged in steady flow).

The impulse-momentum principle: momentum at $t$ + impulse of the external forces = momentum at $t+\Delta t$:

$$\vec H_{AB}+\Delta m\,\vec v_A+\vec F\,\Delta t=\vec H_{AB}+\Delta m\,\vec v_B$$ $$\vec F\,\Delta t=\Delta m(\vec v_B-\vec v_A)$$

Dividing by $\Delta t$ and using $\dot m=\dfrac{dm}{dt}=\rho Av$:

$$\boxed{\vec F=\dot m(\vec v_B-\vec v_A)=\rho Q(\vec v_B-\vec v_A)}$$

where $\vec F$ is the resultant force exerted by the vane on the stream (and $-\vec F$ is the force of the stream on the vane) and $Q=Av$ is the discharge.

• Components: $F_x=\dot m(v_{Bx}-v_{Ax})$, $F_y=\dot m(v_{By}-v_{Ay})$.
• For a vane moving with speed $V$, the same result holds with $\vec v_A,\vec v_B$ taken relative to the vane and $\dot m=\rho A(v_A-V)$ for a single moving vane.