Give step-by-step solution with explanation and final answer:Q-3 For the mechanism shown, below, As and De are fixed joints (fixed pivots);
a) Write the loop closure equations in polar complex form.

Question

Give step-by-step solution with explanation and final answer:Q-3 For the mechanism shown, below, As and De are fixed joints (fixed pivots);
a) Write the loop closure equations in polar complex form. [15 pts]
b) Write the loop-closure equations by separating real and complex parts. [10 pts.]
(ates: BAD is one link (Link-3). For the angular position of link AB: 8a = 21 - (as ~ 81s). Link
DoD is perpendicular to link AD. Therefore, we have: DDo = as ™*** or DsD = as.e*™* 7,
Also, in the figure, points G and H are treated as auxiliary points when going either from As to
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VIPSolutions · Accepted Answer

Let's solve the problem step by step:

---

### **a) Write the loop closure equations in polar complex form.**

#### **Step 1: Assign Vectors and Angles**

Let’s assign the given points and links as vectors in complex form:
- $ A_0 $: origin (0,0)
- $ B $: endpoint of link $ A_0B $ ($ S_{15}, \theta_2 $)
- $ D $: endpoint of link $ A_0D $ ($ S_{36}, \theta_3 $)
- $ D_0 $: fixed point ($ C_1, H $)
- $ C $, $ G $, $ H $: auxiliary points

Let’s define the link lengths and angles as:
- $ \overrightarrow{A_0B} = s_{15} e^{j\theta_2} $
- $ \overrightarrow{BA} = s_{36} e^{j\theta_3} $
- $ \overrightarrow{AD} = a_6 e^{j\theta_6} $
- $ \overrightarrow{DD_0} = a_6 e^{j(\theta_3 + \pi/2)} $ (since $ DD_0 $ is perpendicular to $ AD $)

#### **Step 2: Write Loop Equations**

There are two main loops in the mechanism:

#### **Loop 1: $ A_0 \to B \to D \to A_0 $**
$$
\overrightarrow{A_0B} + \overrightarrow{BD} + \overrightarrow{DA_0} = 0
$$
But $ \overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} $, and $ \overrightarrow{DA_0} = -\overrightarrow{A_0D} $:

So,
$$
s_{15} e^{j\theta_2} + s_{36} e^{j\theta_3} - c_1 - jb_1 = 0
$$
where $ (c_1, b_1) $ are the coordinates of $ D_0 $ from $ A_0 $.

#### **Loop 2: $ D_0 \to D \to A \to A_0 \to D_0 $**
$$
\overrightarrow{D_0D} + \overrightarrow{DA} + \overrightarrow{AA_0} + \overrightarrow{A_0D_0} = 0
$$
But $ \overrightarrow{D_0D} = a_6 e^{j(\theta_3 + \pi/2)} $,
$ \overrightarrow{DA} = -a_6 e^{j\theta_3} $,
$ \overrightarrow{AA_0} = -s_{15} e^{j\theta_2} $,
$ \overrightarrow{A_0D_0} = c_1 + jb_1 $:

So,
$$
a_6 e^{j(\theta_3 + \pi/2)} - a_6 e^{j\theta_3} - s_{15} e^{j\theta_2} + c_1 + jb_1 = 0
$$

---

### **b) Write the loop-closure equations by separating real and complex parts.**

Let’s separate real and imaginary parts for **Loop 1** as an example:

$$
s_{15} e^{j\theta_2} + s_{36} e^{j\theta_3} = c_1 + jb_1
$$

Expanding:
$$
s_{15} (\cos\theta_2 + j\sin\theta_2) + s_{36} (\cos\theta_3 + j\sin\theta_3) = c_1 + jb_1
$$

Separate into real and imaginary parts:

#### **Real part:**
$$
s_{15}\cos\theta_2 + s_{36}\cos\theta_3 = c_1
$$

#### **Imaginary part:**
$$
s_{15}\sin\theta_2 + s_{36}\sin\theta_3 = b_1
$$

Similarly, for **Loop 2**:

Expand:
$$
a_6 e^{j(\theta_3 + \pi/2)} - a_6 e^{j\theta_3} - s_{15} e^{j\theta_2} + c_1 + jb_1 = 0
$$

$ e^{j(\theta_3 + \pi/2)} = \cos(\theta_3 + \pi/2) + j\sin(\theta_3 + \pi/2) = -\sin\theta_3 + j\cos\theta_3 $

So:
$$
a_6(-\sin\theta_3 + j\cos\theta_3) - a_6(\cos\theta_3 + j\sin\theta_3) - s_{15}(\cos\theta_2 + j\sin\theta_2) + c_1 + jb_1 = 0
$$

Group real and imaginary:

**Real:**
$$
-a_6\sin\theta_3 - a_6\cos\theta_3 - s_{15}\cos\theta_2 + c_1 = 0
$$

**Imaginary:**
$$
a_6\cos\theta_3 - a_6\sin\theta_3 - s_{15}\sin\theta_2 + b_1 = 0
$$

---

## **Summary of Answers**

### **a) Loop Closure Equations in Polar Complex Form:**
$$
s_{15} e^{j\theta_2} + s_{36} e^{j\theta_3} = c_1 + jb_1
$$
$$
a_6 e^{j(\theta_3 + \pi/2)} - a_6 e^{j\theta_3} - s_{15} e^{j\theta_2} + c_1 + jb_1 = 0
$$

### **b) Loop Closure Equations (Real and Imaginary Parts):**
**From Loop 1:**
$$
\begin{cases}
s_{15}\cos\theta_2 + s_{36}\cos\theta_3 = c_1 \
s_{15}\sin\theta_2 + s_{36}\sin\theta_3 = b_1 \
\end{cases}
$$

**From Loop 2:**
$$
\begin{cases}
-a_6\sin\theta_3 - a_6\cos\theta_3 - s_{15}\cos\theta_2 + c_1 = 0 \
a_6\cos\theta_3 - a_6\sin\theta_3 - s_{15}\sin\theta_2 + b_1 = 0 \
\end{cases}
$$

---

If you need further explanation for a specific loop or for auxiliary points, let me know!

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