Step 1

We are given a crane boom $OA$ rotating about the $z$ axis with a constantangular velocity.

The angular velocity of the boom $OA$ rotating about the $z$ axis is ${\omega _1} = 0.15\;{\rm{rad/s}}$.

The angular velocity of the boom $OA$ rotating about the $y$ axis is ${\omega _2} = 0.20\;{\rm{rad/s}}$.

We are asked todetermine the velocity and acceleration of point $A$ located at the end of the boom at the instant shown.

 
Step 2

The diagram of the motion of the crane boom is given by,

Using the Pythagoras theorem, the value of $AB$ can be calculated as,

\[\begin{array}{c} OB = \sqrt {{{\left( {OA} \right)}^2} - \left( {A{B^2}} \right)} \\ OB = \sqrt {{{\left( {110\;{\rm{ft}}} \right)}^2} - {{\left( {50\;{\rm{ft}}} \right)}^2}} \\ OB = 97.98\;{\rm{ft}} \end{array}\]
 
Step 3

From the diagram, using the vector addition, the angular velocity of point $A$ is given by,

\[\begin{array}{c} \omega = {\omega _1} + {\omega _2}\\ \omega = \left( {0.2{\bf{j}} + 0.15{\bf{k}}} \right)\;{\rm{rad/s}} \end{array}\]

The direction of rotation of the fixed reference frame coincides with the axis $xyz$ rotating.

The expression for the acceleration of the motor shaft with respect to $xyz$ axis is given by,

\[{\left( {\dot \omega } \right)_{xyz}} = 0\]

The direction of ${\omega _y}$ is along the $y$ axis with respect to the $xyz$ frame. So, the expression for the angular acceleration of the motor shaft is given by,

\[\begin{array}{c} \alpha = \dot \omega \\ \alpha = {\left( {\dot \omega } \right)_{xyz}} + {\omega _1} \times {\omega _2}\\ \alpha = 0 + \left( {0.15{\bf{k}}} \right) \times \left( {0.2{\bf{j}}} \right)\\ \alpha = \left( { - 0.03{\bf{i}}} \right)\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]
 
Step 4

The expression for the position vector of the point $A$ is given by,

\[\begin{array}{c} {{\bf{r}}_A} = \left( {OB{\bf{i}} - AB{\bf{k}}} \right)\\ {{\bf{r}}_A} = \left( {97.98{\bf{i}} - 50{\bf{k}}} \right)\;{\rm{ft}} \end{array}\]

Using the relation between the linear velocity and angular velocity, the expression for the velocity of the point $A$ is given by,

\[\begin{array}{c} {v_A} = \omega \times {{\bf{r}}_A}\\ {v_A} = \left( {0.2{\bf{j}} + 0.15{\bf{k}}} \right)\;{\rm{rad/s}} \times \left( {97.98{\bf{i}} - 50{\bf{k}}} \right)\;{\rm{ft}}\\ {v_A} = \left( { - 19.6{\bf{k}} - 10{\bf{i}} - 0.2{\bf{j}} - 0} \right)\;{\rm{ft/s}}\\ {v_A} = \left( { - 10{\bf{i}} + 14.7{\bf{j}} - 19.6{\bf{k}}} \right)\;{\rm{ft/s}} \end{array}\]
 
Step 5

The expression for the linear acceleration of point $A$ is given by,

\[\begin{array}{c} {a_A} = \alpha \times {{\bf{r}}_A} + \omega \times \left( {{{\bf{v}}_A}} \right)\\ {a_A} = \left[ {\left( { - 0.03{\bf{i}}} \right)\;{\rm{rad/}}{{\rm{s}}^2} \times \left( {97.98{\bf{i}} - 50{\bf{k}}} \right)\;{\rm{ft}}} \right] + \left[ \begin{array}{l} \left( {0.2{\bf{j}} + 0.15{\bf{k}}} \right)\;{\rm{rad/s}} \times \\ \left( { - 10{\bf{i}} + 14.7{\bf{j}} - 19.6{\bf{k}}} \right)\;{\rm{ft/s}} \end{array} \right]\\ {a_A} = \left( {0 + 1.5{\bf{j}} - 2{\bf{k}} + 0 - 3.92{\bf{i}} + 1.5{\bf{j}} - 2.205{\bf{i}} + 0} \right)\\ {a_A} = \left[ { - 6.125{\bf{i}} + 3{\bf{j}} - 2{\bf{k}}} \right]\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]