## Step 1: Free-Body Diagram and Moment Equation **Explanation:** First, isolate the scissor mechanism and consider the force balance. Assume the platform is in static equilibrium. The **weight \( W \) of the platform** acts at the center of gravity \( G \), vertically downward. The hydraulic cylinder \( AB \) applies force \( F_{AB} \) along its length. Other points are pinned or rollers. To find \( F_{AB} \) as a function of \( \theta \), take moments about point \( H \) (since it's a convenient pivot and removes unknown reaction forces at \( H \) from the equation): \[ \sum M_H = \] Let’s denote: - \( W \) = total weight of the platform and load - \( x_G \) = horizontal distance from \( H \) to \( G \) - \( y_G \) = vertical distance from \( H \) to \( G \) The **moment due to weight** about \( H \) is: \[ M_W = W \cdot x_G \] The **moment due to the cylinder force** (projected perpendicular component) is: \[ M_{F_{AB}} = F_{AB} \cdot d_{\perp} \] where \( d_{\perp} \) is the perpendicular distance from \( H \) to the line of action of \( F_{AB} \). **At equilibrium:** \[ F_{AB} \cdot d_{\perp} = W \cdot x_G \] --- ## Step 2: Express \( d_{\perp} \) in Terms of \( \theta \) **Explanation:** Now, express the geometry in terms of \( \theta \): - The scissor mechanism's arms have fixed lengths. - The location of \( A \) and \( B \) changes with \( \theta \). Based on the geometry, **for a typical scissor lift**: - The length \( FD = 54" \) - Angle \( \theta \) is between \( FD \) and the horizontal. The **distance \( d_{\perp} \)** is the perpendicular distance from \( H \) to the line of action of \( AB \). For a standard scissor, this is related to the vertical and horizontal projections and the arm lengths. **For a simplified model:** \[ d_{\perp} = l_{AB} \sin(\phi) \] where \( \phi \) is the angle between \( AB \) and the horizontal (which is also a function of \( \theta \)), and \( l_{AB} \) is the length of cylinder \( AB \). But **generally**, for the scissor mechanism: - \( x_G \) can be written in terms of \( \theta \) from the geometry. - \( d_{\perp} \) is the lever arm of \( AB \) about \( H \), also in terms of \( \theta \). So, \[ F_{AB}(\theta) = \frac{W \cdot x_G(\theta)}{d_{\perp}(\theta)} \] --- ## Final Answer ### **Expression for Cylinder Force** \[ \boxed{ F_{AB}(\theta) = \frac{W \cdot x_G(\theta)}{d_{\perp}(\theta)} } \] where: - \( x_G(\theta) \) is the horizontal distance from \( H \) to \( G \) as a function of \( \theta \) - \( d_{\perp}(\theta) \) is the perpendicular distance from \( H \) to the line of action of \( AB \) as a function of \( \theta \). **To plot \( F_{AB} \) vs. \( \theta \), compute \( x_G \) and \( d_{\perp} \) for \( ^\circ \leq \theta \leq 60^\circ \) based on the given dimensions.** --- **If you want the explicit forms for \( x_G \) and \( d_{\perp} \) in terms of \( \theta \) using the actual numbers, let me know!**

## Step 1: Free-Body Diagram and Moment Equation **Explanation:** We start by analyzing the scissor mechanism and applying the principles of static equilibrium. The total weight \( W \) of the platform acts downward through its center of gravity \( G \). The hydraulic cylinder \( AB \) exerts an upward force \( F_{AB} \). To derive the expression for \( F_{AB} \), we take moments about point \( H \) to eliminate unknown reaction forces at that point: \[ \sum M_H = 0 \] Let’s define: - \( W \) = total weight of the platform and load - \( x_G \) = horizontal distance from \( H \) to \( G \) - \( y_G \) = vertical distance from \( H \) to \( G \) The moment created by the weight about point \( H \) is: \[ M_W = W \cdot x_G \] The moment due to the force from the hydraulic cylinder is: \[ M_{F_{AB}} = F_{AB} \cdot d_{\perp} \] where \( d_{\perp} \) is the perpendicular distance from point \( H \) to the line of action of the force \( F_{AB} \). In equilibrium, we have: \[ F_{AB} \cdot d_{\perp} = W \cdot x_G \] --- ## Step 2: Express \( d_{\perp} \) in Terms of \( \theta \) **Explanation:** Next, we relate the distances to the angle \( \theta \). The distances \( x_G \) and \( d_{\perp} \) depend on the geometry of the scissor mechanism. 1. **Geometry of the Scissor Mechanism:** - Assume the lengths of the scissor arms are fixed, and the angle \( \theta \) varies. - The distance \( x_G \) can be expressed as a function of \( \theta \), typically derived from the geometry of the scissor mechanism. 2. **Perpendicular Distance \( d_{\perp} \):** Using basic trigonometric relationships, we can write: \[ d_{\perp} = l_{AB} \sin(\phi) \] where \( \phi \) is the angle between the hydraulic cylinder \( AB \) and the horizontal line. The angle \( \phi \) is also a function of \( \theta \) due to the scissor mechanism's configuration. Thus, we can express \( F_{AB}(\theta) \) as: \[ F_{AB}(\theta) = \frac{W \cdot x_G(\theta)}{d_{\perp}(\theta)} \] --- ## Final Answer ### **Expression for Cylinder Force** \[ \boxed{ F_{AB}(\theta) = \frac{W \cdot x_G(\theta)}{d_{\perp}(\theta)} } \] #### Where: - \( x_G(\theta) \) is the horizontal distance from \( H \) to \( G \) as a function of \( \theta \). - \( d_{\perp}(\theta) \) is the perpendicular distance from \( H \) to the line of action of \( AB \) as a function of \( \theta \). **To plot \( F_{AB} \) against \( \theta \) for \( 0^\circ < \theta < 60^\circ \), compute specific values for \( x_G \) and \( d_{\perp} \) based on the dimensions provided in the problem.**