Absolutely! Let's break down the problem step by step. --- ## **Step 1: Understanding the Problem** You have four forces acting horizontally on a stage platform. You need to: ### **(a) Find the equivalent force-couple system at O (the origin):** - **Resultant Force \( \vec{R} \)** - **Resultant Moment at O (\( M_O \))** ### **(b) Find the points on the x- and y-axes through which the line of action of the resultant force passes.** --- ## **Step 2: Calculating the Resultant Force \( \vec{R} \)** Let's resolve each force into its x and y components: - \( F_1 = 70\, \text{lb} \) (to the right, along +x) - \( F_2 = 68\, \text{lb} \) (up, along +y) - \( F_3 = 47\, \text{lb} \) (to the left, along -x) - \( F_4 = 71\, \text{lb} \) at an angle \( \theta = 41^\circ \) from the x-axis (top right corner) **Components for \( F_4 \):** - \( F_{4x} = F_4 \cos(41^\circ) \) - \( F_{4y} = F_4 \sin(41^\circ) \) ### **Sum the Forces:** \[ \vec{R}_x = F_1 - F_3 + F_{4x} \] \[ \vec{R}_y = F_2 + F_{4y} \] Plug in the values: - \( F_{4x} = 71 \cos(41^\circ) \approx 53.53 \) lb - \( F_{4y} = 71 \sin(41^\circ) \approx 46.65 \) lb So, \[ \vec{R}_x = 70 - 47 + 53.53 = 76.53\, \text{lb} \] \[ \vec{R}_y = 68 + 46.65 = 114.65\, \text{lb} \] So, the resultant force: \[ \boxed{\vec{R} = 76.53\,\mathbf{i} + 114.65\,\mathbf{j} \ \text{lb}} \] --- ## **Step 3: Calculating the Moment About O** Use \( M_O = \sum (\vec{r} \times \vec{F}) \): - \( F_1 \) acts at O: \( r = 0 \) → moment = 0. - \( F_2 \) acts at A (\( (0, b) \)): - \( M_{O2} = (0 \mathbf{i} + b \mathbf{j}) \times (68 \mathbf{j}) = 0 \) (since both are in y-direction). - \( F_3 \) acts at C (\( (a, 0) \)): - \( M_{O3} = (a \mathbf{i}) \times (-47 \mathbf{i}) = 0 \) (since both are in x-direction). - \( F_4 \) acts at B (\( (a, b) \)), at \( \theta = 41^\circ \): The position vector for \( F_4 \) is \( \vec{r}_4 = a \mathbf{i} + b \mathbf{j} \). The force vector for \( F_4 \) is \( F_4 \cos\theta \mathbf{i} + F_4 \sin\theta \mathbf{j} \). The cross product is: \[ M_{O4} = |a\mathbf{i} + b\mathbf{j}| \times |F_4 \cos\theta \mathbf{i} + F_4 \sin\theta \mathbf{j}| \] \[ = (a \mathbf{i} + b \mathbf{j}) \times (F_4 \cos\theta \mathbf{i} + F_4 \sin\theta \mathbf{j}) \] \[ = a F_4 \sin\theta (\mathbf{i} \times \mathbf{j}) + b F_4 \cos\theta (\mathbf{j} \times \mathbf{i}) \] \[ = a F_4 \sin\theta (\mathbf{k}) - b F_4 \cos\theta (\mathbf{k}) \] \[ = [a F_4 \sin\theta - b F_4 \cos\theta] \mathbf{k} \] **Plug in the numbers:** - \( a = 75 \) in, \( b = 63 \) in - \( F_4 = 71 \) lb, \( \sin(41^\circ) = 0.6561 \), \( \cos(41^\circ) = 0.7547 \) \[ M_{O4} = [75 \times 71 \times 0.6561 - 63 \times 71 \times 0.7547] \] \[ = [75 \times 46.59 - 63 \times 53.59] \] \[ = [3494.25 - 3376.17] \] \[ = 118.08 \ \text{lb-in} \] Now let's check for any other moments: - \( F_2 \)'s moment arm is \( a \) (since it acts at (0, b) and is a force in y at a distance a from O along x): - \( M_{O2} = a \times F_2 \) (clockwise, so negative) - \( M_{O2} = 75 \times 68 = 5100 \) lb-in (CW, so negative) - \( F_3 \) acts at (a, 0), force is -x, arm is -b (since force is down, but arm is from O to a): - \( M_{O3} = -b \times F_3 \) (CCW) - \( M_{O3} = -63 \times 47 = -2961 \) lb-in Summing all the moments (counterclockwise positive): \[ M_O = -5100 + (-2961) + 118.08 = -7942.92 \ \text{lb-in} \] --- ## **Step 4: Line of Action of Resultant Force** The resultant force \( \vec{R} \) can be represented as a single force passing at a certain distance from O, producing the same moment. \[ M_O = R \cdot d \] where \( d \) is the perpendicular distance from O to the line of action of \( \vec{R} \). But, to find the intercepts on the axes: - **x-intercept:** Set \( y = 0 \) and solve for \( x \). - **y-intercept:** Set \( x = 0 \) and solve for \( y \). From the moment equation: \[ M_O = x R_y - y R_x \] **On the x-axis (\( y=0 \)):** \[ M_O = x R_y \implies x = \frac{M_O}{R_y} \] **On the y-axis (\( x=0 \)):** \[ M_O = -y R_x \implies y = -\frac{M_O}{R_x} \] --- ## **Step 5: Plug in the Numbers** - \( R_x = 76.53 \) lb - \( R_y = 114.65 \) lb - \( M_O = -7942.92 \) lb-in **x-intercept:** \[ x = \frac{-7942.92}{114.65} \approx -69.3 \ \text{in} \] **y-intercept:** \[ y = -\frac{-7942.92}{76.53} \approx 103.8 \ \text{in} \] --- # **Summary Table** | Quantity | Value | |--------------------------|-------------------------------| | \( R \) | \( 76.53\,\mathbf{i} + 114.65\,\mathbf{j} \) lb | | \( M_O \) | \( -7942.92 \) lb-in | | x-intercept | \( -69.3 \) in | | y-intercept | \( 103.8 \) in | --- ## **Filling the Answers:** (a) The force-couple system at O: - R = (76.53) i + (114.65) j lb - \( M_O = -7942.92 \) lb-in (b) The line of action of the single resultant R: - On the x-axis: \( x = -69.3 \) in - On the y-axis: \( y = 103.8 \) in --- **Let me know if you want to see the detailed calculations for any particular step!**