Question

Exam duration is 120 minutes and questlons u Return mechanism is shown with dimensions. At the given instant crank AB is on. For a graphical position analysis, draw the mechanism to scale and B and C, and angle θ4 . Then, write down the loop equation of he vector loop shown below and separate the loop equation into en by solving the component equations, calculate L and 04. Do the results 1) In the figure a Quick- measure L. the distance between the mechanism according to t solving the obtained by analytical and graphical methods match ? AC 10 cm 02 2 rad/sec AB 5 cm. 02 30 02 2) Crank AB of the mechanism given in question 1 is rotating CCW with a speed of 2 rad/sed. Calculate the angular velocity of link 4 by a velocity poygon, and state its direction. 3) Differentiate the components of the loop equation of the mechanism given in question1 and calculate or θ4, the angular velocity of link 4 of by solving the derivatives of the component equations. What is the time rate of change of L, the distance between B and Ca

3. Question 1. According to the question

Answer 1

$AB = r cos\theta_2 i+r sin\theta_2 j$

$CB = -l cos\theta_4 i+l sin\theta_4 j$

$di-l cos\theta_4 i+l sin\theta_4 j = r cos\theta_2 i+r sin\theta_2 j$

$di= r cos\theta_2 i+r sin\theta_2 j -(-l cos\theta_4 i+l sin\theta_4 j )$

$di= (r cos\theta_2+l cos\theta_4) i+(r sin\theta_2 -l sin\theta_4) j$

equating i,j components:

$d= r cos\theta_2+l cos\theta_4;r sin\theta_2 -l sin\theta_4 = 0$

with given values:

$10 = 5cos30+l cos\theta_4;5 sin30 -l sin\theta_4 = 0$

$\Rightarrow l cos\theta_4 = 5.67; l sin\theta_4 = 2.5$

$\Rightarrow tan\theta_4 = (2.5/5.67)\Rightarrow \theta_4 = 23.79^0$

$\Rightarrow l*sin23.79 = 2.5\Rightarrow l = 6.2cm$

6.20 10

values match.

2)

16'S 6LT7 3

Angular Velocity of link 4 is:

$w_4= \frac{5.91}{6.2} = 0.953 rad/s CCW$

3)

$d= r cos\theta_2+l cos\theta_4;r sin\theta_2 -l sin\theta_4 = 0$

Differentiating this equations:

$0 =- r sin\theta_2*w_2+\dot l cos\theta_4 - l sin \theta_4*w_4$

$r cos\theta_2*w_2 -\dot l sin\theta_4-l cos\theta_4 *w_4= 0$

substituting given values:

$0 =- 5 sin30*2+\dot l cos23.79 -6.2sin23.79*w_4$

$5 cos30*2 -\dot l sin23.79-6.2 cos23.79 *w_4= 0$

$\Rightarrow 0.915 \dot l -2.5*w_4 = 5$

0.40341 + 5.67:32 ua = 8.66

solving the 2 equations:

$\Rightarrow \dot l = 8.07cm/s;w_4 = 0.953 rad/s$

which is same from the figure.