Certainly! Let’s break down the solution into **two steps** as requested, with explanations at each step, and the **final answer** at the end. --- ## **Step 1: Find the Prying Force \( R \) at C** ### **Given Data:** - Pressure, \( p = 35\, \text{MPa} = 35 \times 10^6\, \text{N/m}^2 \) - Piston radius, \( r = 50\, \text{mm} = .05\, \text{m} \) #### **Force Developed by the Piston:** \[ F_{\text{piston}} = p \cdot A = p \cdot \pi r^2 \] \[ F_{\text{piston}} = 35 \times 10^6 \cdot \pi \cdot (.05)^2 \] \[ F_{\text{piston}} = 35 \times 10^6 \cdot \pi \cdot .0025 \] \[ F_{\text{piston}} = 35 \times 10^6 \cdot .00785 \] \[ F_{\text{piston}} = 274,890\, \text{N} \] **Explanation:** We first calculate the force output of the hydraulic piston using the given pressure and the piston area. --- ## **Step 2: Express the Prying Force \( R \) in Terms of the Jaw Angle \(\theta\)** Assume the mechanism is a simple lever (see the jaw's geometry) with the following dimensions (from the figure): - Distance from C to B = 110 mm - Distance from A to C = 120 mm - Distance from C to the line of action of F (piston) = 110 mm Let’s apply the principle of moments about point C for equilibrium: \[ F_{\text{piston}} \cdot d_{\text{F}} = R \cdot d_{\text{R}} \] Where: - \( d_{\text{F}} \) is the perpendicular distance from C to the line of action of the piston force. - \( d_{\text{R}} \) is the perpendicular distance from C to the line of action of the prying force. From the geometry (for the given angle \(\theta\)), the distances change, but for \(\theta = 45^\circ\) as asked: \[ d_{\text{F}} = \text{distance from C to the piston line} = 110\, \text{mm} \] \[ d_{\text{R}} = \text{perpendicular distance from C to the line of action of R} \] Assuming symmetric jaws, \( d_{\text{R}} = 120 \sin\theta \) mm. So, \[ R = \frac{F_{\text{piston}} \cdot d_{\text{F}}}{d_{\text{R}}} \] \[ R = \frac{274,890 \cdot .11}{.12 \sin\theta} \] \[ R = \frac{30,238}{.12 \sin\theta} \] \[ R = \frac{252,000}{\sin\theta} \text{ N} \] **Explanation:** We use the moment equilibrium about point C to relate the piston force to the jaw prying force \( R \), expressing \( R \) as a function of the jaw angle \(\theta\). --- ## **Final Answer** ### **Prying force at \(\theta = 45^\circ\):** \[ R_{45^\circ} = \frac{252,000}{\sin 45^\circ} = \frac{252,000}{.7071} \approx 356,400\, \text{N} \] **Minimum value of \( R \):** \( R \) is minimized when \(\sin\theta\) is maximized, i.e., at \(\theta = 90^\circ\): \[ R_{\text{min}} = \frac{252,000}{\sin 90^\circ} = 252,000\, \text{N} \] --- ### **Summary Table** | \(\theta\) | \(\sin\theta\) | \(R\) (N) | |------------|----------------|---------------| | \(45^\circ\) | .7071 | 356,400 | | \(90^\circ\) | 1.000 | 252,000 | - **Minimum value of \( R \) is \(252,000\, \text{N}\) at \(\theta = 90^\circ\).** --- If you need the other forces (e.g., in the links), let me know!