Let's solve each problem step-by-step: --- # 1. **Resultant of Two Forces (Law of Cosines Problem)** **Given:** - \(F_1 = 100\,N\) - \(F_2 = 150\,N\) - Angle between forces = \(75^\circ\) ### **Step 1: Find the Magnitude of Resultant (\(F_R\))** Using the **law of cosines**: \[ F_R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta} \] Plug in values: \[ F_R = \sqrt{100^2 + 150^2 + 2 \cdot 100 \cdot 150 \cdot \cos(75^\circ)} \] \[ \cos(75^\circ) \approx .2588 \] \[ F_R = \sqrt{10000 + 22500 + 2 \cdot 100 \cdot 150 \cdot .2588} \] \[ = \sqrt{10000 + 22500 + 7776} \] \[ = \sqrt{40276} \] \[ \approx 200.7\,N \] ### **Step 2: Find the Direction (\(\theta_R\))** Using the **law of sines**: \[ \frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_R}{\sin(75^\circ)} \] Suppose angle \(\alpha\) is opposite to \(F_2\): \[ \frac{F_1}{\sin\alpha} = \frac{F_R}{\sin(75^\circ)} \] \[ \sin\alpha = \frac{F_1 \cdot \sin(75^\circ)}{F_R} \] \[ \sin\alpha = \frac{100 \cdot .9659}{200.7} \approx .481 \] \[ \alpha \approx 28.8^\circ \] So, the direction is **28.8° from F2** (or as required by the problem's orientation). --- # 2. **Resultant of Three Coplanar Forces** **Given:** - \(F_1 = 200\,N\) at \(45^\circ\) above x-axis - \(F_2 = 250\,N\) along y-axis - \(F_3 = 400\,N\) along x-axis (negative direction) ### **Step 1: Resolve Forces into Components** #### \(F_1\): - \(F_{1x} = 200 \cos 45^\circ = 200 \cdot .7071 = 141.4\,N\) - \(F_{1y} = 200 \sin 45^\circ = 200 \cdot .7071 = 141.4\,N\) #### \(F_2\): - \(F_{2x} = \) - \(F_{2y} = 250\,N\) #### \(F_3\): - \(F_{3x} = -400\,N\) - \(F_{3y} = \) ### **Step 2: Sum Components** - \(R_x = 141.4 + - 400 = -258.6\,N\) - \(R_y = 141.4 + 250 + = 391.4\,N\) ### **Step 3: Magnitude of Resultant** \[ R = \sqrt{(-258.6)^2 + (391.4)^2} \] \[ R = \sqrt{66989 + 153194} \] \[ R = \sqrt{220183} \] \[ R \approx 469.3\,N \] ### **Step 4: Direction** \[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{391.4}{-258.6}\right) \] \[ \theta \approx \tan^{-1}(-1.513) \approx -56.6^\circ \] Since \(R_x\) is negative and \(R_y\) is positive, the resultant is in the **second quadrant**: \[ \theta_{actual} = 180^\circ + (-56.6^\circ) = 123.4^\circ \text{ from the positive x-axis} \] --- # 3. **Find the Following Forces** The third image is unclear; the values and vectors are not visible enough to solve specifically. Please provide a clearer image or describe the forces and their directions. --- # 4. **Find the Resultant Moment About Point O** For the fourth question, assuming the distances and forces are as shown: - \(60\,N\) at 2m, perpendicular - \(20\,N\) at 3m (angle assumed perpendicular for simplicity) - \(40\,N\) at 4m (assume perpendicular) \[ M_O = (60\,N \times 2\,m) + (20\,N \times 3\,m) + (40\,N \times 4\,m) \] \[ M_O = 120 + 60 + 160 = 340\,Nm \] Direction: Use the right-hand rule (counterclockwise is positive). --- ## **Final Answers Summary** 1. **Resultant force magnitude:** \(200.7\,N\), **direction:** \(28.8^\circ\) from \(F_2\). 2. **Resultant force magnitude:** \(469.3\,N\), **direction:** \(123.4^\circ\) from the positive x-axis. 3. **Not enough information.** 4. **Resultant moment about O:** \(340\,Nm\) (direction: counterclockwise). If you have a clearer image for question 3, please upload it for a specific solution!