Step 1

We are given the system with various dimensions.

We are asked to determine the velocity and acceleration of the slider block C.

 
Step 2

The free body diagram of the bar AB is shown as:

We have the length of bar AB is ${r_{AB}} = 0.5\;{\rm{m}}$.

We have the angle of bar AB with horizontal is ${\theta _1} = 45^\circ $.

We have the angular velocity of bar AB is ${\omega _{AB}} = 4\;{\rm{rad/s}}$.

We have the angular acceleration of bar AB is ${\alpha _{AB}} = 6\;{\rm{rad/}}{{\rm{s}}^2}$.

We have the vector form of angular acceleration of bar AB is $\overrightarrow {{\alpha _{AB}}} = 6k\;{\rm{rad/}}{{\rm{s}}^2}$.

 
Step 3

The formula to calculate the velocity of link AB is,

\[{v_B} = {\omega _{AB}} \times {r_{AB}}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} {v_B} = \left( {4\;{\rm{rad/s}}} \right)\left( {0.5\;{\rm{m}}} \right)\\ {v_B} = 2\;{\rm{m/s}} \end{array}\]
 
Step 5

The expression to calculate the vector form of the length of bar AB is,

\[\overrightarrow {{r_{AB}}} = \left( {{r_{AB}}\cos {\theta _1}} \right)i + \left( {{r_{AB}}\sin {\theta _1}} \right)j\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{l} \overrightarrow {{r_{AB}}} = \left( {0.5\cos 45^\circ \;{\rm{m}}} \right)i + \left( {0.5\sin 45^\circ \;{\rm{m}}} \right)j\\ \overrightarrow {{r_{AB}}} = \left( {0.35i + 0.35j} \right)\;{\rm{m}} \end{array}\]
 
Step 7

The formula to calculate the acceleration of link AB is,

\[\overrightarrow {{a_B}} = \left( {\overrightarrow {{\alpha _{AB}}} \times \overrightarrow {{r_{AB}}} } \right) - {\left( {{\omega _{AB}}} \right)^2}\overrightarrow {{r_{AB}}} \]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{l} \overrightarrow {{a_B}} = \left\{ {\left( {6k\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {0.35i + 0.35j} \right)\;{\rm{m}}} \right\} - \left\{ {{{\left( {4\;{\rm{rad/s}}} \right)}^2}\left( {0.35i + 0.35j} \right)\;{\rm{m}}} \right\}\\ \overrightarrow {{a_B}} = \left( { - 5.5\sqrt 2 i - 2.5\sqrt 2 j} \right)\;{\rm{m/}}{{\rm{s}}^2} \end{array}\]
 
Step 9

The free body diagram of the bar BC is shown as:

We have the length of bar BC is ${r_{BC}} = 1\;{\rm{m}}$.

We have the angle of bar BC from horizontal is ${\theta _2} = 60^\circ $.

We have the angle of bar BC from vertical is ${\theta _3} = 30^\circ $.

 
Step 10

The formula to calculate the distance between point B and instantaneous center by triangular relation is,

\[\frac{{{r_{B/IC}}}}{{\sin {\theta _3}}} = \frac{{{r_{BC}}}}{{\sin {\theta _1}}}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{c} \frac{{{r_{B/IC}}}}{{\sin 30^\circ }} = \frac{{1\;{\rm{m}}}}{{\sin 45^\circ }}\\ {r_{B/IC}} = 0.71\;{\rm{m}} \end{array}\]
 
Step 12

The formula to calculate the distance between point C and instantaneous center by triangular relation is,

\[\frac{{{r_{C/IC}}}}{{\sin \left( {{\theta _1} + {\theta _2}} \right)}} = \frac{{{r_{BC}}}}{{\sin {\theta _1}}}\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{c} \frac{{{r_{C/IC}}}}{{\sin \left( {45 + 60} \right)^\circ }} = \frac{{1\;{\rm{m}}}}{{\sin 45^\circ }}\\ \frac{{{r_{C/IC}}}}{{\sin 105^\circ }} = \frac{{1\;{\rm{m}}}}{{\sin 45^\circ }}\\ {r_{C/IC}} = 1.36\;{\rm{m}} \end{array}\]
 
Step 14

The formula to calculate the angular velocity of bar BC is,

\[{v_B} = {\omega _{BC}}{r_{B/IC}}\]
 
Step 15

Substitute the values in the above expression.

\[\begin{array}{c} \left( {2\;{\rm{m/s}}} \right) = {\omega _{BC}}\left( {0.71\;{\rm{m}}} \right)\\ {\omega _{BC}} = 2.82\;{\rm{rad/s}} \end{array}\]
 
Step 16

The formula to calculate the velocity of slider C is,

\[{v_C} = {\omega _{BC}}{r_{B/IC}}\]
 
Step 17

Substitute the values in the above expression.

\[\begin{array}{c} {v_C} = \left( {2.82\;{\rm{rad/s}}} \right)\left( {1.37\;{\rm{m}}} \right)\\ {v_C} = 3.86\;{\rm{m/s}} \end{array}\]
 
Step 18

The free body diagram of bar BC for relative acceleration is shown as:

 
Step 19

The expression to calculate the vector form of the length of bar BC is,

\[\overrightarrow {{r_{BC}}} = \left( {{r_{BC}}\cos {\theta _2}} \right)i - \left( {{r_{BC}}\sin {\theta _2}} \right)j\]
 
Step 20

Substitute the values in the above expression.

\[\begin{array}{l} \overrightarrow {{r_{BC}}} = \left( {1\cos 60^\circ \;{\rm{m}}} \right)i - \left( {1\sin 60^\circ \;{\rm{m}}} \right)j\\ \overrightarrow {{r_{BC}}} = \left( {0.5i - 0.86j} \right)\;{\rm{m}} \end{array}\]
 
Step 21

The formula to calculate the acceleration of link BC is,

\[\overrightarrow {{a_C}} = \overrightarrow {{a_B}} + \left( {\overrightarrow {{\alpha _{BC}}} \times \overrightarrow {{r_{BC}}} } \right) - {\left( {{\omega _{BC}}} \right)^2}\overrightarrow {{r_{BC}}} \]
 
Step 22

Substitute the values in the above expression.

\[\begin{array}{l} - {a_C}i = \left\{ \begin{array}{l} \left[ {\left( { - 5.5\sqrt 2 i - 2.5\sqrt 2 j} \right)\;{\rm{m/}}{{\rm{s}}^2}} \right] + \left\{ {\left( {{\alpha _{BC}}k} \right) \times \left( {0.5i - 0.86j} \right)\;{\rm{m}}} \right\}\\ - \left\{ {{{\left( {2.82\;{\rm{rad/s}}} \right)}^2}\left( {0.5i - 0.86j} \right)\;{\rm{m}}} \right\} \end{array} \right\}\\ - {a_C}i = \left\{ \begin{array}{l} \left[ {\left( { - 7.778{\bf{i}} - 3.53{\bf{j}}} \right)\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right]\\ + \left[ {\left\{ {\left( {0.5{\alpha _{BC}}} \right){\bf{j}} + \left( {0.86{\alpha _{BC}}} \right){\bf{i}}} \right\}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right]\\ - \left[ {\left( {3.97{\bf{i}}} \right)\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right] + \left[ {\left( {6.83{\bf{j}}} \right){\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right] \end{array} \right\}\\ - {a_C}i = \left[ {\left\{ {\left( { - 11.748} \right) + \left( {0.86{\alpha _{BC}}} \right)} \right\}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right]{\bf{i}} + \left[ {\left( {3.3} \right) + \left( {0.5{\alpha _{BC}}} \right)\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right]{\bf{j}} \end{array}\]
 
Step 23

On comparing $j$ components in the above expression, we get:

\[\begin{array}{c} \left( {3.3 + 0.5{\alpha _{BC}}} \right) = 0\\ {\alpha _{BC}} = - 6.6\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]
 
Step 24

On comparing $i$ components in the above expression, we get:

\[ - {a_C} = \left[ {\left\{ {\left( { - 11.748} \right) + \left( {0.86{\alpha _{BC}}} \right)} \right\}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right]\]
 
Step 25

Substitute the values in the above expression.

\[\begin{array}{c} - {a_C} = \left[ {\left\{ {\left( { - 11.748} \right) + \left( {0.86\left( { - 6.6} \right)} \right)} \right\}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right]\\ {a_C} = 17.42\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]