Let's solve **Question 1** step by step: #### **Given:** - **F₁** = 20 lbs, **Direction:** 120° 30' 50" (Quadrant II) - **F₂** = 2 kips (1 kip = 1000 lbs), **m = 2/3**, **Q = II** *(This is ambiguous; let's assume the direction is similar to F₁)* - **F₃** = 300,000 dynes, **Direction:** 60° 10' 10" North of East - **F₄** = mass = 20 slugs, **Direction:** 340° 20' 20" - **Required:** Find the resultant force in Newtons and its direction. --- ### **Step 1: Convert All Forces to Newtons** **1 lb = 4.4482 N** **1 kip = 1000 lbs = 4448.2 N** **1 dyne = 1e-5 N** **Weight (F = mg), 1 slug = 32.174 lbs = 14.5939 kg** --- #### **F₁: 20 lbs** \[ F_1 = 20 \times 4.4482 = 88.964~N \] Direction: \(120^\circ 30' 50''\) --- #### **F₂: 2 kips** \[ F_2 = 2 \times 1000 \times 4.4482 = 2 \times 4448.2 = 8896.4~N \] --- #### **F₃: 300,000 dynes** \[ F_3 = 300,000 \times 1 \times 10^{-5} = 3~N \] --- #### **F₄: Weight of 20 slugs** \[ F_4 = 20 \text{ slugs} \times 32.174~\text{lb/slug} \times 4.4482~\text{N/lb} \] First, 20 slugs to lbs: \[ 20 \times 32.174 = 643.48~\text{lbs} \] Now, lbs to Newtons: \[ 643.48 \times 4.4482 = 2861.3~N \] --- ### **Step 2: Break Each Force into Components** Let’s denote: - \(F_{ix} = F_i \cos \theta_i\) - \(F_{iy} = F_i \sin \theta_i\) #### **Convert Angles to Decimal Degrees** - \(120^\circ 30' 50'' = 120 + 30/60 + 50/3600 \approx 120.514^\circ\) - \(60^\circ 10' 10'' = 60 + 10/60 + 10/3600 \approx 60.169^\circ\) - \(340^\circ 20' 20'' = 340 + 20/60 + 20/3600 \approx 340.339^\circ\) --- ### **Assume F₂ is in the same direction as F₁ (for simplicity, since only magnitude is given).** --- ### **Calculate Components** #### **F₁:** \[ F_{1x} = 88.964 \cos(120.514^\circ) = 88.964 \times (-0.502) = -44.67~N \] \[ F_{1y} = 88.964 \sin(120.514^\circ) = 88.964 \times 0.865 = 76.93~N \] --- #### **F₂:** \[ F_{2x} = 8896.4 \cos(120.514^\circ) = 8896.4 \times (-0.502) = -4466.88~N \] \[ F_{2y} = 8896.4 \sin(120.514^\circ) = 8896.4 \times 0.865 = 7695.58~N \] --- #### **F₃:** \[ F_{3x} = 3 \cos(60.169^\circ) = 3 \times 0.495 = 1.485~N \] \[ F_{3y} = 3 \sin(60.169^\circ) = 3 \times 0.869 = 2.607~N \] --- #### **F₄:** \[ F_{4x} = 2861.3 \cos(340.339^\circ) = 2861.3 \times 0.939 = 2687.1~N \] \[ F_{4y} = 2861.3 \sin(340.339^\circ) = 2861.3 \times (-0.344) = -984.3~N \] --- ### **Step 3: Add Components** #### **Sum of X-components:** \[ F_{Rx} = (-44.67) + (-4466.88) + 1.485 + 2687.1 = -1823.0~N \] #### **Sum of Y-components:** \[ F_{Ry} = 76.93 + 7695.58 + 2.607 + (-984.3) = 6790.8~N \] --- ### **Step 4: Magnitude of Resultant** \[ F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} \] \[ F_R = \sqrt{(-1823.0)^2 + (6790.8)^2} \] \[ F_R = \sqrt{3323705 + 46117418} \] \[ F_R = \sqrt{49441123} \] \[ F_R \approx 7031~N \] --- ### **Step 5: Direction of Resultant** \[ \theta = \arctan\left(\frac{F_{Ry}}{F_{Rx}}\right) \] \[ \theta = \arctan\left(\frac{6790.8}{-1823.0}\right) \] \[ \theta \approx \arctan(-3.726) \] \[ \theta \approx -75.1^\circ \] Since \(F_{Rx}\) is negative and \(F_{Ry}\) is positive, the angle is in the second quadrant, so: \[ \theta_{actual} = 180^\circ - 75.1^\circ = 104.9^\circ \] --- ## **Final Answer:** \[ \boxed{ \text{Resultant force} = 7031~\text{N} \text{ at } 104.9^\circ \text{ from the positive x-axis} } \] --- **Let me know if you want to proceed with the next question or any specific one!**