Let's solve the problem step by step: ### **Given Data** - **Weight of rods, \( W \) = \( 2.0 \times 10^{-2} \) lb** - **Each rod length, \( L_0 = 12 \) in = 1 ft** - **Water temperature = \( 100^\circ F \)** - **Surface tension of water at \( 100^\circ F \): \( \sigma \) (from tables)** - At \( 100^\circ F \), \( \sigma \approx 0.066 \) lb/ft ### **Step 1: Understanding the Forces** The vertical force \( P \) needed to pull the rods free from the surface is the sum of: - The weight of the rods \( W \) - The vertical component of the surface tension forces acting along the rods Since the equilateral triangle has three sides, three rods are in contact with the water. The surface tension acts tangentially to the water surface at the contact points, and only the vertical component resists the upward pull. ### **Step 2: Calculating the Total Surface Tension Force** For each rod, the surface tension force is: \[ F_{\text{st, rod}} = \sigma \times L_0 \] There are 3 rods, so the total surface tension force is: \[ F_{\text{st, total}} = 3 \times (\sigma \times L_0) \] The force acts tangent to the water surface, but the vertical component (which resists pulling) is the full force since the rods are tangent at the surface. ### **Step 3: Total Vertical Force Needed** \[ P = W + F_{\text{st, total}} \] ### **Step 4: Substitute the Values** - \( W = 2.0 \times 10^{-2} \) lb - \( \sigma = 0.066 \) lb/ft - \( L_0 = 1 \) ft \[ F_{\text{st, total}} = 3 \times 0.066 \times 1 = 0.198 \text{ lb} \] \[ P = 0.02 + 0.198 = 0.218 \text{ lb} \] ### **Step 5: Final Answer (4 significant figures)** \[ \boxed{P = 0.2180 \ \text{lb}} \] --- ## **Summary Table** | Quantity | Value | Units | |---------------------------|------------|----------| | Weight, \( W \) | 0.020 | lb | | Surface tension, \( \sigma \) | 0.066 | lb/ft | | Rod length, \( L_0 \) | 1.0 | ft | | Surface tension force | 0.198 | lb | | **Total force, \( P \)** | **0.2180** | **lb** | --- **Final Answer:** \[ \boxed{P = 0.2180 \ \text{lb}} \] Let me know if you need a breakdown of the units or a diagram!

Let's revisit the problem with the corrected surface tension value at 100°F: --- ### **Given Data (Updated)** - **Weight of the rods, \( W = 2.0 \times 10^{-2} \) lb** - **Length of each rod, \( L_0 = 12 \) in = 1 ft** - **Temperature = 100°F** - **Surface tension of water at 100°F, \( \sigma = 4.79 \times 10^{-3} \) lb/ft** --- ### **Step 1: Understand the forces** The total force \( P \) needed to detach the rods from the water surface has two components: - The weight of the rods, \( W \) - The vertical component of the surface tension along the contact line Since the rods are tangent to the water surface at the contact points, the surface tension acts along the contact line, providing a force resisting separation. --- ### **Step 2: Determine the surface tension force** The surface tension force along each rod is: \[ F_{\text{st, per rod}} = \sigma \times L_0 \] Given: \[ \sigma = 4.79 \times 10^{-3} \ \text{lb/ft} \] \[ L_0 = 1 \ \text{ft} \] Thus: \[ F_{\text{st, per rod}} = 4.79 \times 10^{-3} \times 1 = 4.79 \times 10^{-3} \ \text{lb} \] --- ### **Step 3: Calculate the total surface tension force** There are 3 rods, so: \[ F_{\text{st, total}} = 3 \times 4.79 \times 10^{-3} = 0.01437 \ \text{lb} \] --- ### **Step 4: Calculate total force \( P \)** Total force needed is: \[ P = W + F_{\text{st, total}} \] \[ P = 0.02 + 0.01437 = 0.03437 \ \text{lb} \] ### **Step 5: Final answer (rounded to 4 significant figures)** \[ \boxed{P = 0.03437 \ \text{lb}} \] --- ### **Summary:** | Quantity | Value | Units | |---------------------------|---------------------------|----------| | Weight of rods, \( W \) | 0.020 | lb | | Surface tension, \( \sigma \) | 4.79 × 10\(^{-3}\) | lb/ft | | Length of each rod, \( L_0 \) | 1.0 | ft | | Total surface tension force | 0.01437 | lb | | **Total force \( P \)** | **0.03437** | lb | --- **Final answer:** \[ \boxed{P = \mathbf{0.03437 \, \text{lb}}} \] Let me know if you'd like further clarification!

Let's analyze the problem step-by-step to estimate the precision in the calculated flow rate based on measurement uncertainties. --- ### **Given Data:** - **Measured mass flow rate:** \( \dot{m} = 0.2 \) kg/s - **Scale reading accuracy:** ± 0.05 kg - **Stopwatch accuracy:** ± 0.2 s - **Time intervals:** - (a) 10 seconds - (b) 60 seconds (1 minute) --- ### **Objective:** Estimate the **uncertainty** (or the **least count**) in the **calculated flow rate** \( \dot{m} \) for each time interval, considering uncertainties in mass measurement and time measurement. --- ### **Step 1: Understand how flow rate is calculated** Flow rate: \[ \dot{m} = \frac{\text{Mass collected } (m)}{\text{Time } (t)} \] - The mass \( m \) is measured using a scale. - The time \( t \) is measured using a stopwatch. --- ### **Step 2: Determine uncertainties** - **Mass measurement uncertainty:** \[ \Delta m = \pm 0.05 \ \text{kg} \] - **Time measurement uncertainty:** \[ \Delta t = \pm 0.2 \ \text{s} \] --- ### **Step 3: Propagate uncertainties to flow rate** The flow rate is: \[ \dot{m} = \frac{m}{t} \] The **uncertainty** in \( \dot{m} \), denoted \( \Delta \dot{m} \), can be approximated using the **error propagation formula**: \[ \left( \frac{\Delta \dot{m}}{\dot{m}} \right)^2 = \left( \frac{\Delta m}{m} \right)^2 + \left( \frac{\Delta t}{t} \right)^2 \] --- ### **Step 4: Calculate uncertainties for each case** #### (a) **Time interval = 10 s** - Measured mass: \[ m = \dot{m} \times t = 0.2 \times 10 = 2.0 \ \text{kg} \] - Relative uncertainty in mass: \[ \frac{\Delta m}{m} = \frac{0.05}{2.0} = 0.025 \] - Relative uncertainty in time: \[ \frac{\Delta t}{t} = \frac{0.2}{10} = 0.02 \] - Total relative uncertainty: \[ \left( \frac{\Delta \dot{m}}{\dot{m}} \right) = \sqrt{(0.025)^2 + (0.02)^2} = \sqrt{0.000625 + 0.0004} = \sqrt{0.001025} \approx 0.032 \] - Therefore, the **absolute uncertainty** in the flow rate: \[ \Delta \dot{m} = \dot{m} \times 0.032 = 0.2 \times 0.032 = 0.0064 \ \text{kg/s} \] --- #### (b) **Time interval = 60 s** - Measured mass: \[ m = 0.2 \times 60 = 12.0 \ \text{kg} \] - Relative uncertainty in mass: \[ \frac{\Delta m}{m} = \frac{0.05}{12} \approx 0.00417 \] - Relative uncertainty in time: \[ \frac{\Delta t}{t} = \frac{0.2}{60} \approx 0.00333 \] - Total relative uncertainty: \[ \sqrt{(0.00417)^2 + (0.00333)^2} \approx \sqrt{0.0000174 + 0.0000111} \approx \sqrt{0.0000285} \approx 0.00534 \] - Absolute uncertainty in flow rate: \[ \Delta \dot{m} = 0.2 \times 0.00534 \approx 0.00107 \ \text{kg/s} \] --- ### **### Final Results:** | Time Interval | Estimated Flow Rate \( \dot{m} \) | Uncertainty \( \Delta \dot{m} \) | |----------------|----------------------------------|------------------------------| | 10 seconds | **0.2 kg/s** | **± 0.0064 kg/s** | | 1 minute (60s) | **0.2 kg/s** | **± 0.00107 kg/s** | --- ### **Summary:** - For a 10-second interval, the flow rate can be determined with approximately **± 0.0064 kg/s** uncertainty. - For a 60-second interval, the uncertainty reduces to about **± 0.00107 kg/s** due to longer measurement time improving precision. --- **Note:** Longer measurement periods yield better precision because the relative error in time measurement diminishes, and the fixed mass measurement error becomes less significant relative to the total mass. --- Let me know if you'd like any further clarification or details!

### **Step 1. Governing relation** The mass flow rate is measured as $$ \dot m = \frac{\Delta m}{\Delta t} $$ where * $\Delta m$ = collected mass, * $\Delta t$ = time interval. The **relative uncertainty** in $\dot m$ is obtained by uncertainty propagation: $$ \left(\frac{u_{\dot m}}{\dot m}\right)^2 = \left(\frac{u_m}{\Delta m}\right)^2 + \left(\frac{u_t}{\Delta t}\right)^2 $$ where $u_m$ = uncertainty in collected mass, $u_t$ = uncertainty in time. --- ### **Step 2. Instrument uncertainties** * Scale reading to nearest 0.05 kg → **use half the least count**: $$ u_m = \frac{0.05}{2} = 0.025\ \text{kg} $$ * Stopwatch accurate to 0.2 s → **half the least count**: $$ u_t = \frac{0.2}{2} = 0.1\ \text{s} $$ --- ### **Step 3. Case (a): $\Delta t = 10\ \text{s}$** * Mass collected: $$ \Delta m = \dot m \, \Delta t = 0.2 \times 10 = 2.0\ \text{kg} $$ * Relative uncertainty in $\Delta m$: $$ \frac{u_m}{\Delta m} = \frac{0.025}{2.0} = 0.0125 = 1.25\% $$ * Relative uncertainty in $\Delta t$: $$ \frac{u_t}{\Delta t} = \frac{0.1}{10} = 0.01 = 1.00\% $$ * Combined uncertainty: $$ \frac{u_{\dot m}}{\dot m} = \sqrt{(0.0125)^2 + (0.01)^2} = \sqrt{0.00015625 + 0.0001} = \sqrt{0.00025625} \approx 0.0160 = 1.60\% $$ * Absolute uncertainty: $$ u_{\dot m} = 0.0160 \times 0.2 = 0.0032\ \text{kg/s} $$ **Final result:** $$ \boxed{ \dot m = 0.200 \pm 0.0032\ \text{kg/s} \quad (\text{about } 1.60\%) } $$ --- ### **Step 4. Case (b): $\Delta t = 60\ \text{s}$** * Mass collected: $$ \Delta m = 0.2 \times 60 = 12.0\ \text{kg} $$ * Relative uncertainty in $\Delta m$: $$ \frac{u_m}{\Delta m} = \frac{0.025}{12} \approx 0.00208 = 0.208\% $$ * Relative uncertainty in $\Delta t$: $$ \frac{u_t}{\Delta t} = \frac{0.1}{60} \approx 0.00167 = 0.167\% $$ * Combined uncertainty: $$ \frac{u_{\dot m}}{\dot m} = \sqrt{(0.00208)^2 + (0.00167)^2} \approx \sqrt{4.33 \times 10^{-6} + 2.79 \times 10^{-6}} = \sqrt{7.12 \times 10^{-6}} \approx 0.00267 = 0.267\% $$ * Absolute uncertainty: $$ u_{\dot m} = 0.00267 \times 0.2 = 0.00053\ \text{kg/s} $$ **Final result:** $$ \boxed{ \dot m = 0.200 \pm 0.00053\ \text{kg/s} \quad (\text{about } 0.267\%) } $$ --- ### **✅ Final Summary** - **For a 10-s interval:** approximately **±1.60%** uncertainty in the flow rate. - **For a 60-s interval:** approximately **±0.267%** uncertainty. Longer measurement intervals reduce the relative uncertainty due to the fixed instrument errors.

### **Step 1. Governing relation** The mass flow rate is measured as $$ \dot{m} = \frac{\Delta m}{\Delta t} $$ where * $\Delta m$ = collected mass * $\Delta t$ = time interval The **relative uncertainty** in $\dot{m}$ is obtained by uncertainty propagation: $$ \left(\frac{u_{\dot{m}}}{\dot{m}}\right)^2 = \left(\frac{u_m}{\Delta m}\right)^2 + \left(\frac{u_t}{\Delta t}\right)^2 $$ where $u_m$ = uncertainty in collected mass, and $u_t$ = uncertainty in time. --- ### **Step 2. Instrument uncertainties** * Scale reading to nearest 0.05 kg → **use half the least count**: $$ u_m = \frac{0.05}{2} = 0.025\ \text{kg} $$ * Stopwatch accurate to 0.2 s → **half the least count**: $$ u_t = \frac{0.2}{2} = 0.1\ \text{s} $$ --- ### **Step 3. Case (a): $\Delta t = 10\ \text{s}$** * Mass collected: $$ \Delta m = \dot{m} \times \Delta t = 0.2 \times 10 = 2.0\ \text{kg} $$ * Relative uncertainty in $\Delta m$: $$ \frac{u_m}{\Delta m} = \frac{0.025}{2.0} = 0.0125 \quad (1.25\%) $$ * Relative uncertainty in $\Delta t$: $$ \frac{u_t}{\Delta t} = \frac{0.1}{10} = 0.01 \quad (1.00\%) $$ * Combined relative uncertainty: $$ \frac{u_{\dot{m}}}{\dot{m}} = \sqrt{(0.0125)^2 + (0.01)^2} = \sqrt{0.00015625 + 0.0001} = \sqrt{0.00025625} \approx 0.0160 \quad (1.60\%) $$ * Absolute uncertainty: $$ u_{\dot{m}} = 0.0160 \times 0.2 = 0.0032\ \text{kg/s} $$ **Final result:** $$ \boxed{ \dot{m} = 0.200 \pm 0.0032\ \text{kg/s} \quad (\text{about } 1.60\%) } $$ --- ### **Step 4. Case (b): $\Delta t = 60\ \text{s}$** * Mass collected: $$ \Delta m = 0.2 \times 60 = 12.0\ \text{kg} $$ * Relative uncertainty in $\Delta m$: $$ \frac{u_m}{\Delta m} = \frac{0.025}{12} \approx 0.00208 \quad (0.208\%) $$ * Relative uncertainty in $\Delta t$: $$ \frac{u_t}{\Delta t} = \frac{0.1}{60} \approx 0.00167 \quad (0.167\%) $$ * Combined relative uncertainty: $$ \frac{u_{\dot{m}}}{\dot{m}} = \sqrt{(0.00208)^2 + (0.00167)^2} \approx \sqrt{4.33 \times 10^{-6} + 2.79 \times 10^{-6}} = \sqrt{7.12 \times 10^{-6}} \approx 0.00267 \quad (0.267\%) $$ * Absolute uncertainty: $$ u_{\dot{m}} = 0.00267 \times 0.2 = 0.00053\ \text{kg/s} $$ **Final result:** $$ \boxed{ \dot{m} = 0.200 \pm 0.00053\ \text{kg/s} \quad (\text{about } 0.267\%) } $$ --- ### **✅ Final Summary** - **For a 10-s interval:** approximately **±1.60%** uncertainty in the flow rate. - **For a 60-s interval:** approximately **±0.267%** uncertainty. Longer measurement durations reduce the relative uncertainty due to the fixed instrument errors.

# Solution (Formal, Fully Worked) **Given (from figure):** Three horizontal layers with thicknesses and specific gravities: - **Top layer:** - Thickness \( h_1 = 12.0\ \text{ft} \) - Specific gravity \( S_1 = 0.6 \) - **Middle layer:** - Thickness \( h_2 = 15.0\ \text{ft} \) - Specific gravity \( S_2 = 1.2 \) - **Bottom layer:** - Thickness \( h_3 = 8.0\ \text{ft} \) - Specific gravity \( S_3 = 2.8 \) Assume the water/air pressure at the top surface (point A) is taken as zero gauge. Use the unit weight of water: $$ \gamma_w = 62.4\ \frac{\text{lb}}{\text{ft}^3} $$ Then the unit weight of each layer is: $$ \gamma_i = S_i \cdot \gamma_w $$ ### **Step 1: Compute Unit Weights** - For layer 1: $$ \gamma_1 = 0.6 \cdot 62.4 = 37.44\ \frac{\text{lb}}{\text{ft}^3} $$ - For layer 2: $$ \gamma_2 = 1.2 \cdot 62.4 = 74.88\ \frac{\text{lb}}{\text{ft}^3} $$ - For layer 3: $$ \gamma_3 = 2.8 \cdot 62.4 = 174.72\ \frac{\text{lb}}{\text{ft}^3} $$ ### **Step 2: Depths of Points** Depths of points measured from top surface: - \( z_A = 0\ \text{ft} \) - \( z_B = h_1 = 12\ \text{ft} \) - \( z_C = h_1 + h_2 = 27\ \text{ft} \) - \( z_D = h_1 + h_2 + h_3 = 35\ \text{ft} \) ### **Step 3: Compute Vertical Pressure** The vertical pressure (gauge) at depth \( z \) is the weight of overlying material: $$ p(z) = \sum_{i\ \text{above } z} \gamma_i h_i $$ ### **Step 4: Calculate Pressures** 1. **Point A (surface)**: $$ p_A = 0\ \text{psf} \quad (0.000\ \text{ksf},\; 0.000\ \text{psi}) $$ 2. **Point B (at 12 ft)**: $$ p_B = \gamma_1 h_1 = 37.44 \cdot 12 = 449.28\ \text{psf} = 0.4493\ \text{ksf} = \frac{449.28}{144} = 3.120\ \text{psi} $$ 3. **Point C (at 27 ft)**: $$ p_C = \gamma_1 h_1 + \gamma_2 h_2 = 449.28 + 74.88 \cdot 15 = 449.28 + 1123.20 = 1572.48\ \text{psf} = 1.5725\ \text{ksf} = \frac{1572.48}{144} = 10.918\ \text{psi} $$ 4. **Point D (at 35 ft, base)**: $$ p_D = p_C + \gamma_3 h_3 = 1572.48 + 174.72 \cdot 8 = 1572.48 + 1397.76 = 2970.24\ \text{psf} = 2.9702\ \text{ksf} = \frac{2970.24}{144} = 20.626\ \text{psi} $$ --- ## Final Values (Rounded Reasonably) - \( p_A = 0\ \text{psf} \quad (=0.000\ \text{ksf},\ 0.000\ \text{psi}) \) - \( p_B = 449.3\ \text{psf} \quad (=0.4493\ \text{ksf},\ 3.120\ \text{psi}) \) - \( p_C = 1572.5\ \text{psf} \quad (=1.5725\ \text{ksf},\ 10.918\ \text{psi}) \) - \( p_D = 2970.2\ \text{psf} \quad (=2.9702\ \text{ksf},\ 20.626\ \text{psi}) \) **Note:** Values given are gauge pressures due to the overburden. If absolute pressures are required, add atmospheric pressure to each gauge pressure.