Let's analyze the step-by-step--- ## ** Data:** Vector | Magn | Direction/ | |------------------------|-----------------| **A | 30 | 60° | | **B** | 15 lb | 25° | | **C** | 10 lb | 20° | | **E** | ? | 40° (direction from the tip of A) | | **F** | ? | ? (direction from the tip of C) | --- ## **Objective:** - To find the magnitudes of vectors **E** and **F** using the component method, given the vector relation: \[ A + B - C = E - F \] which can be rewritten as: \[ A + B - C - E + F = \] or equivalently: \[ A + B - C = E - F \] --- ## **Step 1: Express all vectors in components** ### **a) Vector A** - Magnitude = 30 lb - Direction = 60° Components: \[ A_x = 30 \cos 60° = 30 \times .5 = 15 \] \[ A_y = 30 \sin 60° = 30 \times \frac{\sqrt{3}}{2} \approx 30 \times .866 = 25.98 \] ### **b) Vector B** - Magnitude = 15 lb - Direction = 25° Components: \[ B_x = 15 \cos 25° \approx 15 \times .9063 \approx 13.59 \] \[ B_y = 15 \sin 25° \approx 15 \times .4226 \approx 6.34 \] ### **c) Vector C** - Magnitude = 10 lb - Direction = 20° Components: \[ C_x = 10 \cos 20° \approx 10 \times .9397 \approx 9.40 \] \[ C_y = 10 \sin 20° \approx 10 \times .342 \approx 3.42 \] --- ## **Step 2: Calculate \(A + B - C\) in components** \[ \text{Sum} = (A_x + B_x - C_x, A_y + B_y - C_y) \] \[ x\text{-component}: \] \[ 15 + 13.59 - 9.40 = 19.19 \] \[ y\text{-component}: \] \[ 25.98 + 6.34 - 3.42 = 28.90 \] --- ## **Step 3: Find the magnitude of \(A + B - C\)** \[ |A + B - C| = \sqrt{(19.19)^2 + (28.90)^2} \] \[ = \sqrt{368.39 + 835.21} = \sqrt{1203.60} \approx 34.66 \text{ lb} \] --- ## **Step 4: Determine \(E\) and \(F\)** Given: \(A + B - C = E - F\) - Both \(E\) and \(F\) are vectors. - \(E\) is at 40°, with unknown magnitude \(E\). - \(F\) has unknown magnitude \(F\) and direction (unknown). Since the problem states the relation and gives the directions for \(E\) and \(F\), and the directions are relative, we proceed assuming: \[ E = E \times (\cos 40°, \sin 40°) \] \[ F = F \times (\cos \theta_F, \sin \theta_F) \] But from the diagram, \(F\) is at 20°, so: \[ F = F \times (\cos 20°, \sin 20°) \] --- ## **Step 5: Express \(E\) and \(F\) in components** \[ E_x = E \cos 40° \approx E \times .766 \] \[ E_y = E \sin 40° \approx E \times .643 \] \[ F_x = F \cos 20° \approx F \times .9397 \] \[ F_y = F \sin 20° \approx F \times .342 \] --- ## **Step 6: Write the component equation** From the relation: \[ A + B - C = E - F \] or: \[ \Rightarrow (A + B - C)_{x,y} = (E_x - F_x, E_y - F_y) \] which yields: \[ 19.19 = E \times .766 - F \times .9397 \] \[ 28.90 = E \times .643 - F \times .342 \] --- ## **Step 7: Solve the system of equations** Expressed as: \[ \begin{cases} 19.19 = .766 E - .9397 F \\ 28.90 = .643 E - .342 F \end{cases} \] Multiply the first equation by .643 and the second by .766 to eliminate \(E\): \[ (19.19)(.643) = (.766)(.643) E - (.9397)(.643) F \] \[ (28.90)(.766) = (.643)(.766) E - (.342)(.766) F \] Calculations: \[ 12.34 = .493 E - .6048 F \] \[ 22.16 = .493 E - .2622 F \] Subtract the first from the second: \[ (22.16 - 12.34) = (.493 E - .493 E) - (.2622 F - .6048 F) \] \[ 9.82 = + .3426 F \] Solve for \(F\): \[ F = \frac{9.82}{.3426} \approx 28.66 \text{ lb} \] Now, substitute \(F\) back into the first equation: \[ 19.19 = .766 E - .9397 \times 28.66 \] \[ 19.19 = .766 E - 26.96 \] \[ .766 E = 19.19 + 26.96 = 46.15 \] \[ E = \frac{46.15}{.766} \approx 60.22 \text{ lb} \] --- ## **## **Final Results:** | Quantity | Magnitude | |------------|--------------| | **E** | **60.22 lb** | | **F** | **28.66 lb** | --- ## **Summary:** - The magnitude of **E** is approximately **60.22 lb**. - The magnitude of **F** is approximately **28.66 lb**. This complete solution uses component method to resolve the vectors and solve the system of equations.