Let's solve the problem step by step as a helpful tutor: --- ## **Given Data** - **Link lengths:** \( L_2 = a = 25 \) in \( L_3 = b = 78 \) in \( \text{Offset} = L_4 = c = 8 \) in - **Angles and velocities:** \( \theta_2 = 55^\circ \) \( \omega_2 = -11 \) rad/s (CW, so negative) \( \alpha_2 = 15 \) rad/s² \( \theta_3 = 170.7^\circ \) - **Find:** **Acceleration of the slider** (\(\ddot{d}\)) in in/s². --- ## **Step 1: Position Analysis** From the diagram and the loop-closure equation for the crank-slider: \[ a (\cos \theta_2, \sin \theta_2) + b (\cos \theta_3, \sin \theta_3) = (d, c) \] This gives two equations: 1. \( a \cos\theta_2 + b\cos\theta_3 = d \) 2. \( a \sin\theta_2 + b\sin\theta_3 = c \) We can use these for velocity and acceleration analysis. --- ## **Step 2: Velocity Analysis** Differentiate the position equations with respect to time: ### \( x \)-direction: \[ -a \sin\theta_2 \cdot \omega_2 - b \sin\theta_3 \cdot \omega_3 = \dot{d} \] ### \( y \)-direction: \[ a \cos\theta_2 \cdot \omega_2 + b \cos\theta_3 \cdot \omega_3 = 0 \] Solve for \(\omega_3\): \[ a \cos\theta_2 \cdot \omega_2 + b \cos\theta_3 \cdot \omega_3 = 0 \implies \omega_3 = - \frac{a \cos\theta_2}{b \cos\theta_3} \omega_2 \] --- ## **Step 3: Acceleration Analysis** Differentiate the velocity equations with respect to time: ### \( x \)-direction: \[ - a \cos\theta_2 \cdot \omega_2^2 - a \sin\theta_2 \cdot \alpha_2 - b \cos\theta_3 \cdot \omega_3^2 - b \sin\theta_3 \cdot \alpha_3 = \ddot{d} \] ### \( y \)-direction: \[ - a \sin\theta_2 \cdot \omega_2^2 + a \cos\theta_2 \cdot \alpha_2 - b \sin\theta_3 \cdot \omega_3^2 + b \cos\theta_3 \cdot \alpha_3 = 0 \] --- ## **Step 4: Plug in Known Values** ### Convert angles to radians: \[ \theta_2 = 55^\circ = 0.9599~\text{rad} \] \[ \theta_3 = 170.7^\circ = 2.9806~\text{rad} \] ### Calculate cosines and sines: \[ \begin{align*} \cos\theta_2 &= \cos(55^\circ) = 0.5736 \\ \sin\theta_2 &= \sin(55^\circ) = 0.8192 \\ \cos\theta_3 &= \cos(170.7^\circ) = -0.9845 \\ \sin\theta_3 &= \sin(170.7^\circ) = 0.1755 \\ \end{align*} \] --- ## **Step 5: Find \( \omega_3 \) (Angular Velocity of Link 3)** \[ \omega_3 = - \frac{a \cos\theta_2}{b \cos\theta_3} \omega_2 \] \[ \omega_3 = - \frac{25 \times 0.5736}{78 \times (-0.9845)} \times (-11) \] \[ \omega_3 = - \frac{14.34}{-76.791} \times (-11) \] \[ \omega_3 = 0.1867 \times 11 = 2.05~\text{rad/s} \] --- ## **Step 6: Find \( \alpha_3 \) (Angular Acceleration of Link 3)** Use the \( y \)-direction acceleration equation: \[ - a \sin\theta_2 \omega_2^2 + a \cos\theta_2 \alpha_2 - b \sin\theta_3 \omega_3^2 + b \cos\theta_3 \alpha_3 = 0 \] Plug in numbers: - \( a = 25 \) - \( b = 78 \) - \( \sin\theta_2 = 0.8192 \) - \( \cos\theta_2 = 0.5736 \) - \( \omega_2 = -11 \) - \( \alpha_2 = 15 \) - \( \sin\theta_3 = 0.1755 \) - \( \cos\theta_3 = -0.9845 \) - \( \omega_3 = 2.05 \) Calculate each term: - \( -a \sin\theta_2 \omega_2^2 = -25 \times 0.8192 \times 121 = -2478.88 \) - \( a \cos\theta_2 \alpha_2 = 25 \times 0.5736 \times 15 = 215.1 \) - \( -b \sin\theta_3 \omega_3^2 = -78 \times 0.1755 \times (2.05^2) = -78 \times 0.1755 \times 4.2025 = -78 \times 0.7375 = -57.53 \) - \( b \cos\theta_3 \alpha_3 = 78 \times (-0.9845) \times \alpha_3 = -76.791 \alpha_3 \) Plug into equation: \[ -2478.88 + 215.1 - 57.53 - 76.791 \alpha_3 = 0 \] \[ -2321.31 - 76.791 \alpha_3 = 0 \] \[ -76.791 \alpha_3 = 2321.31 \] \[ \alpha_3 = \frac{2321.31}{-76.791} = -30.22~\text{rad/s}^2 \] --- ## **Step 7: Find Slider Acceleration (\(\ddot{d}\))** Use the \( x \)-direction acceleration equation: \[ \ddot{d} = - a \cos\theta_2 \omega_2^2 - a \sin\theta_2 \alpha_2 - b \cos\theta_3 \omega_3^2 - b \sin\theta_3 \alpha_3 \] Plug in values: - \( -a \cos\theta_2 \omega_2^2 = -25 \times 0.5736 \times 121 = -1734.51 \) - \( -a \sin\theta_2 \alpha_2 = -25 \times 0.8192 \times 15 = -307.20 \) - \( -b \cos\theta_3 \omega_3^2 = -78 \times (-0.9845) \times (2.05^2) = -78 \times (-0.9845) \times 4.2025 = 78 \times 0.9845 \times 4.2025 = 323.37 \) - \( -b \sin\theta_3 \alpha_3 = -78 \times 0.1755 \times (-30.22) = 78 \times 0.1755 \times 30.22 = 414.20 \) Sum up: \[ \ddot{d} = -1734.51 - 307.20 + 323.37 + 414.20 \] \[ \ddot{d} = (-1734.51 - 307.20) + (323.37 + 414.20) \] \[ \ddot{d} = -2041.71 + 737.57 = -1304.14~\text{in/s}^2 \] --- ## **Final Answer** \[ \boxed{\ddot{d} = -1304.14~\text{in/s}^2} \] --- ### **Summary Table** | Step | Value | Formula Used | |------------------|---------------|--------------| | \(\omega_3\) | 2.05 rad/s | \(- \frac{a \cos\theta_2}{b \cos\theta_3} \omega_2\) | | \(\alpha_3\) | -30.22 rad/s² | Acceleration loop, y-direction | | \(\ddot{d}\) | -1304.14 in/s²| Acceleration loop, x-direction | --- ### **Explanation** - Negative sign means the slider is accelerating to the left (opposite to the positive direction defined in the diagram). - All computations are shown step by step. - Let me know if you want more explanation about any step!