1. Calculate the resultant of the forces for each of the following. **a)** Forces: 3 N left, 5 N right. - Step 1: State the problem: Find the resultant force when 3 N acts to the left and 5 N acts to the right. - Step 2: Use the formula for forces in opposite directions: $$F_{res} = F_{right} - F_{left}$$ - Step 3: Calculate: $$F_{res} = 5 - 3 = 2\,\text{N to the right}$$ **b)** Forces: 4 N left, 8 N right. - Step 1: State the problem: Find the resultant force when 4 N acts to the left and 8 N acts to the right. - Step 2: Use the formula: $$F_{res} = F_{right} - F_{left}$$ - Step 3: Calculate: $$F_{res} = 8 - 4 = 4\,\text{N to the right}$$ **c)** Forces: 6 N left, 4 N right, 3 N up, 3 N down. - Step 1: Resolve horizontal forces: $$F_{res,x} = 4 - 6 = -2\,\text{N (left)}$$ - Step 2: Resolve vertical forces: $$F_{res,y} = 3 - 3 = 0\,\text{N}$$ - Step 3: Since vertical forces cancel, resultant force is horizontal: $$2\,\text{N to the left}$$ **d)** Forces: 3 N left, 7 N right, 3 N up, 3 N down, 4 N right. - Step 1: Sum horizontal forces: $$F_{res,x} = (7 + 4) - 3 = 11 - 3 = 8\,\text{N to the right}$$ - Step 2: Sum vertical forces: $$F_{res,y} = 3 - 3 = 0\,\text{N}$$ - Step 3: Resultant force is horizontal: $$8\,\text{N to the right}$$ 2. Calculate the resultant of the forces for each of the following. **a)** Forces: 8 N right, 6 N down. - Step 1: State the problem: Find the resultant force when 8 N acts right and 6 N acts down. - Step 2: Use Pythagoras theorem for perpendicular forces: $$F_{res} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\,\text{N}$$ - Step 3: Direction angle $$\theta = \tan^{-1}\left(\frac{6}{8}\right) = \tan^{-1}(0.75) \approx 36.87^\circ \text{ below horizontal right}$$ **b)** Forces: 4 N right, 5 N right, 12 N down. - Step 1: Sum horizontal forces: $$F_{res,x} = 4 + 5 = 9\,\text{N right}$$ - Step 2: Vertical force: $$F_{res,y} = 12\,\text{N down}$$ - Step 3: Resultant magnitude: $$F_{res} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\,\text{N}$$ - Step 4: Direction angle: $$\theta = \tan^{-1}\left(\frac{12}{9}\right) = \tan^{-1}(1.333) \approx 53.13^\circ \text{ below horizontal right}$$ **c)** Forces: 12 N up, 5 N right. - Step 1: Resultant magnitude: $$F_{res} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\,\text{N}$$ - Step 2: Direction angle: $$\theta = \tan^{-1}\left(\frac{5}{12}\right) = \tan^{-1}(0.4167) \approx 22.62^\circ \text{ above horizontal right}$$ **d)** Forces: 15 N up, 25 N right. - Step 1: Resultant magnitude: $$F_{res} = \sqrt{15^2 + 25^2} = \sqrt{225 + 625} = \sqrt{850} \approx 29.15\,\text{N}$$ - Step 2: Direction angle: $$\theta = \tan^{-1}\left(\frac{25}{15}\right) = \tan^{-1}(1.6667) \approx 59.04^\circ \text{ above horizontal right}$$