Kinematics And Dynamics Of Machine Martin Solution 44

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Kinematics and Dynamics of Machine Martin Solution 44

Kinematics and dynamics of machine martin is a textbook that covers the theory and applications of mechanical engineering. It provides a comprehensive introduction to the kinematics and dynamics of mechanisms, machines, and systems. It also includes numerous examples and problems to help students master the concepts and skills.

One of the problems in the textbook is problem 44, which involves finding the angular velocity and acceleration of a four-bar linkage. This problem can be challenging for some students, but with the right approach and tools, it can be solved easily. In this article, we will show you how to solve problem 44 in kinematics and dynamics of machine martin using a step-by-step guide.

Step 1: Draw the four-bar linkage and label the dimensions

The first step is to draw the four-bar linkage and label the dimensions. The four-bar linkage consists of four rigid links connected by four revolute joints. The link lengths are given as L1 = 0.3 m, L2 = 0.4 m, L3 = 0.5 m, and L4 = 0.6 m. The input link is link 2, which rotates at a constant angular velocity of ω2 = 10 rad/s counterclockwise. The output link is link 4, which rotates about point D. The fixed link is link 1, which connects points A and B. The coupler link is link 3, which connects points C and D.

Here is a diagram of the four-bar linkage with the dimensions labeled:

Step 2: Apply the loop-closure equation

The second step is to apply the loop-closure equation, which relates the position vectors of the links in a closed loop. The loop-closure equation for the four-bar linkage is given by:

$\vec{r}_{12} + \vec{r}_{23} + \vec{r}_{34} + \vec{r}_{41} = \vec{0}$

where $\vec{r}_{ij}$ is the position vector from point i to point j. Using the complex number notation, we can write the position vectors as:

$\vec{r}_{12} = L_1 e^{j\theta_1}$

$\vec{r}_{23} = L_2 e^{j\theta_2}$

$\vec{r}_{34} = L_3 e^{j\theta_3}$

$\vec{r}_{41} = -L_4 e^{j\theta_4}$

where $\theta_i$ is the angle between link i and the horizontal axis, and $j$ is the imaginary unit. Substituting these expressions into the loop-closure equation, we get:

$L_1 e^{j\theta_1} + L_2 e^{j\theta_2} + L_3 e^{j\theta_3} - L_4 e^{j\theta_4} = 0$

This equation can be simplified by dividing both sides by $L_1 e^{j\theta_1}$, which gives:

$1 + \frac{L_2}{L_1} e^{j(\theta_2 - \theta_1)} + \frac{L_3}{L_1} e^{j(\theta_3 - \theta_1)} - \frac{L_4}{L_1} e^{j(\theta_4 - \theta_1)} = 0$

This equation can be further simplified by defining some constants as follows:

$k_{12} = \frac{L_2}{L_1}, k_{13} = \frac{L_3}{L_1}, k_{14} = \frac{L_4}{L_1}, \phi_{12} = \theta_2 - \theta_1, \phi_{13} = \theta_3 - \theta_1, \phi_{14} = \theta_4 - \theta_1$

Using these constants, we can rewrite the loop-closure equation as: 4aad9cdaf3