Let's solve **Question 1** step by step: ### **Given Forces:** 1. \( F_1 = 20 \) lbs, \( -120^\circ 30' 50'' \) 2. \( F_2 = 2 \) kips, \( m = 2/3, Q = II \) 3. \( F_3 = 300,000 \) dynes, \( 60^\circ 10' 10'' \) North of East 4. \( F_4 = \) mass \( = 20 \) slugs, \( 340^\circ 20' 20'' \) Let's break down each part: --- #### **Step 1: Convert All Forces to Newtons** - **1 lbf (pound-force) = 4.4482 N** - **1 kip = 1000 lbf = 4448.2 N** - **1 dyne = \( 10^{-5} \) N** - **Weight (force) = mass (in slugs) × g (32.174 ft/s² = 9.80665 m/s²)** --- #### **F1:** - \( F_1 = 20 \) lbs \( = 20 \times 4.4482 = 88.964 \) N --- #### **F2:** - \( F_2 = 2 \) kips \( = 2 \times 4448.2 = 8896.4 \) N - "m = 2/3, Q = II" — This means the angle with respect to x-axis: - Slope \( m = \tan{\theta} = 2/3 \implies \theta = \arctan(2/3) \approx 33.69^\circ \) - Quadrant II: \( \theta = 180^\circ - 33.69^\circ = 146.31^\circ \) --- #### **F3:** - \( F_3 = 300,000 \) dynes \( = 300,000 \times 10^{-5} = 3 \) N - \( 60^\circ 10' 10'' \) North of East \(= 60.1694^\circ \) --- #### **F4:** - Mass = 20 slugs, Weight = \( 20 \times 32.174 = 643.48 \) lbf \( = 643.48 \times 4.4482 = 2862.9 \) N - Angle: \( 340^\circ 20' 20'' = 340.3389^\circ \) --- ### **Step 2: Resolve All Forces into x- and y-components** #### **Convert angles to degrees:** - \( F_1 \): \( -120^\circ 30' 50'' = -120.514^\circ \) - \( F_2 \): \( 146.31^\circ \) - \( F_3 \): \( 60.1694^\circ \) - \( F_4 \): \( 340.3389^\circ \) #### **Components:** Let \( F_x = F \cos\theta \), \( F_y = F \sin\theta \) --- #### **F1:** \( F_{1x} = 88.964 \cos(-120.514^\circ) \) \( F_{1y} = 88.964 \sin(-120.514^\circ) \) Calculate: - \( \cos(-120.514^\circ) = -0.502 \) - \( \sin(-120.514^\circ) = -0.865 \) So, - \( F_{1x} = 88.964 \times (-0.502) = -44.68 \) N - \( F_{1y} = 88.964 \times (-0.865) = -76.93 \) N --- #### **F2:** \( F_{2x} = 8896.4 \cos(146.31^\circ) \) \( F_{2y} = 8896.4 \sin(146.31^\circ) \) - \( \cos(146.31^\circ) = -0.832 \) - \( \sin(146.31^\circ) = 0.554 \) So, - \( F_{2x} = 8896.4 \times (-0.832) = -7404.0 \) N - \( F_{2y} = 8896.4 \times (0.554) = 4928.0 \) N --- #### **F3:** \( F_{3x} = 3 \cos(60.1694^\circ) \) \( F_{3y} = 3 \sin(60.1694^\circ) \) - \( \cos(60.1694^\circ) = 0.495 \) - \( \sin(60.1694^\circ) = 0.869 \) So, - \( F_{3x} = 3 \times 0.495 = 1.485 \) N - \( F_{3y} = 3 \times 0.869 = 2.607 \) N --- #### **F4:** \( F_{4x} = 2862.9 \cos(340.3389^\circ) \) \( F_{4y} = 2862.9 \sin(340.3389^\circ) \) - \( \cos(340.3389^\circ) = 0.939 \) - \( \sin(340.3389^\circ) = -0.344 \) So, - \( F_{4x} = 2862.9 \times 0.939 = 2688.6 \) N - \( F_{4y} = 2862.9 \times (-0.344) = -985.8 \) N --- ### **Step 3: Add All x and y Components** Sum x-components: \[ F_{x,\text{total}} = -44.68 + (-7404.0) + 1.485 + 2688.6 = -4758.6 \text{ N} \] Sum y-components: \[ F_{y,\text{total}} = -76.93 + 4928.0 + 2.607 + (-985.8) = 3867.9 \text{ N} \] --- ### **Step 4: Compute Resultant Magnitude and Direction** \[ F_R = \sqrt{F_{x,\text{total}}^2 + F_{y,\text{total}}^2} \] \[ F_R = \sqrt{(-4758.6)^2 + (3867.9)^2} \] \[ F_R = \sqrt{22646917 + 14963791} \] \[ F_R = \sqrt{37610708} \approx 6133 \text{ N} \] Direction (relative to x-axis): \[ \theta = \arctan\left(\frac{F_{y,\text{total}}}{F_{x,\text{total}}}\right) \] \[ \theta = \arctan\left(\frac{3867.9}{-4758.6}\right) = \arctan(-0.813)\approx -39.1^\circ \] But since \( F_x \) is negative and \( F_y \) is positive, the angle is in the **second quadrant**: \[ \theta = 180^\circ - 39.1^\circ = 140.9^\circ \] --- ## **Final Answer:** **1. The resultant force is approximately \( \boxed{6133 \text{ N}} \) at an angle of \( \boxed{140.9^\circ} \) (measured counterclockwise from the positive x-axis).** --- Let me know if you want step-by-step solutions for the other questions!

Certainly! Let's proceed with **Question 2**: --- ## **Question 2:** *In the figure, two boxes of identical masses (50 kg each) experience a sliding frictional force with coefficient \( \mu = 0.25 \).* ### **Part (a): Find the acceleration of the two boxes.** ### **Part (b): Determine the tension in the cord connecting the two boxes.** --- ## **Step 1: Understand the problem** - Both boxes have mass \( m = 50\,kg \). - Coefficient of kinetic friction \( \mu = 0.25 \). - The boxes are connected by a cord over a pulley (assumed), and a force \( P \) is applied (or perhaps gravity component). *Since the figure isn't provided here, typically, such problems involve pulling or inclined planes.* **Assumption:** Based on typical problems, let's assume: - The boxes are on an inclined plane at an angle \( \theta \) (say 30°, as indicated in the diagram). - The force \( P \) is applied horizontally or along the incline. - The problem involves calculating acceleration considering friction and possibly tension in the cord. --- ## **Step 2: Calculate the forces** ### **a) Frictional force:** \[ f_{fric} = \mu N \] Where \( N \) is the normal force. On an inclined plane of angle \( \theta \): \[ N = m g \cos \theta \] \[ f_{fric} = \mu m g \cos \theta \] --- ### **b) Components of gravity along the incline:** \[ F_{gravity} = m g \sin \theta \] --- ## **Step 3: Write equations of motion** Assuming no other external forces, the net force along the incline: \[ F_{net} = T - f_{fric} - m g \sin \theta \] But if the boxes are connected with a cord and pulled, the tension \( T \) acts to accelerate the system. **If a force \( P \) is applied**, we need to consider its direction. For simplicity, let's assume the boxes are pulled along the incline by an external force \( P \). --- ## **Part (a): Acceleration** ### **Case 1: No external force other than gravity and friction** The net force on each box: \[ F_{net} = m a = m g \sin \theta - \mu m g \cos \theta \] \[ a = g (\sin \theta - \mu \cos \theta) \] Plugging in values: \[ g = 9.8\,m/s^2, \quad \theta = 30^\circ \] \[ a = 9.8 (\sin 30^\circ - 0.25 \cos 30^\circ) \] Calculate: - \( \sin 30^\circ = 0.5 \) - \( \cos 30^\circ \approx 0.866 \) \[ a = 9.8 (0.5 - 0.25 \times 0.866) = 9.8 (0.5 - 0.2165) = 9.8 \times 0.2835 \approx 2.78\,m/s^2 \] **So, the acceleration of each box is approximately \( \boxed{2.78\,m/s^2} \).** --- ## **Part (b): Tension in the cord** - For the **entire system**, total mass: \[ M_{total} = 2 \times 50\,kg = 100\,kg \] - Net force causing acceleration: \[ F_{net} = M_{total} \times a = 100 \times 2.78 = 278\,N \] - The tension \( T \) in the cord can be found from one of the boxes: \[ T = m (a + g \sin \theta) + f_{fric} \] But more straightforwardly, for each box: \[ T = m a + f_{fric} + m g \sin \theta \] Alternatively, because the boxes are connected and accelerating together: \[ T = m (a + g \sin \theta) + f_{fric} \] Calculating: \[ T = 50 (2.78 + 9.8 \times 0.5) + 0.25 \times 50 \times 9.8 \times 0.866 \] Compute each term: - \( 9.8 \times 0.5 = 4.9 \) - \( 50 \times (2.78 + 4.9) = 50 \times 7.68 = 384\,N \) - Friction force: \[ f_{fric} = 0.25 \times 50 \times 9.8 \times 0.866 \approx 0.25 \times 50 \times 8.477 \approx 0.25 \times 423.85 \approx 105.96\,N \] Add to the tension: \[ T = 384 + 105.96 \approx 490\,N \] **Final tension in the cord: approximately \( \boxed{490\,N} \).** --- ## **Summary:** - **(a)** The acceleration of each box is approximately **2.78 m/s²**. - **(b)** The tension in the cord is approximately **490 N**. --- *Note:* These calculations are based on typical assumptions. If the actual problem involves different forces or configurations, please specify or provide additional details for precise solutions. --- Would you like to proceed with **Question 3** next?