Step 1

We are given the time of lift is $t = 5\;{\rm{s}}$ and the angular velocity of the motor shaft is $\omega = 100\left( {4 + t} \right)\;{\rm{rad/s}}$.

We are asked to determine the distance the load W is lifted.

 
Step 2

The diagram of the system is shown as:

We have the radius of gear A is ${r_A} = 40\;{\rm{mm}}$.

We have the radius of gear B is ${r_B} = 225\;{\rm{mm}}$.

We have the radius of gear C is ${r_C} = 30\;{\rm{mm}}$.

We have the radius of gear D is ${r_D} = 300\;{\rm{mm}}$.

We have the radius of gear E is ${r_E} = 50\;{\rm{mm}}$.

 
Step 3

The formula to calculate the angular displacement of the gear A at $t = 5\;{\rm{s}}$ is,

\[\int\limits_0^{{\theta _A}} {d\theta } = \int\limits_0^t {\omega dt} \]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \int\limits_0^{{\theta _A}} {d\theta } = \int\limits_0^5 {100\left( {4 + t} \right)dt} \\ \left[ \theta \right]_0^{{\theta _A}} = 100\left[ {4t + \frac{{{t^2}}}{2}} \right]_0^5\\ \left[ {{\theta _A} - 0} \right] = 100\left[ {4\left( 5 \right) + \frac{{{{\left( 5 \right)}^2}}}{2} - 0} \right]\\ {\theta _A} = 3250\;{\rm{rad}} \end{array}\]
 
Step 5

The formula to calculate the angular displacement of the gear B is,

\[\frac{{{\theta _B}}}{{{\theta _A}}} = \frac{{{r_A}}}{{{r_B}}}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} \frac{{{\theta _B}}}{{3250\;{\rm{rad}}}} = \frac{{40\;{\rm{mm}}}}{{225\;{\rm{mm}}}}\\ {\theta _B} = 577.78\;{\rm{rad}} \end{array}\]

The gear C is attached on the same shaft as gear B. i.e., ${\theta _C} = {\theta _B} = 577.78\;{\rm{rad}}$.

 
Step 7

The formula to calculate the angular displacement of the gear D is,

\[\frac{{{\theta _D}}}{{{\theta _C}}} = \frac{{{r_C}}}{{{r_D}}}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} \frac{{{\theta _D}}}{{577.78\;{\rm{rad}}}} = \frac{{30\;{\rm{mm}}}}{{300\;{\rm{mm}}}}\\ {\theta _D} = 57.78\;{\rm{rad}} \end{array}\]

The gear E is attached on the same shaft as gear D. i.e., ${\theta _E} = {\theta _D} = 57.78\;{\rm{rad}}$.

 
Step 9

The formula to calculate the distance at which the load W is lifted is,

\[s = {r_E}{\theta _E}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{l} s = \left( {50\;{\rm{mm}} \times \frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right)\left( {57.78\;{\rm{rad}}} \right)\\ s = 2.90\;{\rm{m}} \end{array}\]