Let us represent

\(\displaystyle{T}_{{1}}\) as magnitude of tension in thecord between the drum and the pulley

the net torque is calculated accordingto the equation

\(\displaystyle\sum{T}={I}\alpha\)

so the net torque exerted on the drum will be

\(\displaystyle\sum{T}={I}_{{1}}\alpha_{{1}}\)

Where \(\displaystyle{I}_{{1}}\) is the moment of inertia of the drum

\(\displaystyle\alpha_{{1}}\) is the angular acceleration of the drum

if we take that the cable does not slip then we can write

\(\displaystyle-{T}_{{1}}{r}_{{1}}+{T}={\left({m}_{{1}}{{r}_{{1}}^{{2}}}\right)}{\left({\frac{{{a}}}{{{r}_{{1}}}}}\right)}\)

in the above equation

T is the counter clockwise torque providedby the motor and a is the acceleration of the cord as from the given we can see that

\(\displaystyle{a}={1.3}\frac{{m}}{{s}^{{2}}}\)

in the next step we take \(\displaystyle{T}_{{2}}\) as magnitude of tension in the cordbetween the crate and the pulley and \(\displaystyle{I}_{{2}}\) as the moment of inertia of the pulley next we apply newtons second law of motion to thepulley then we get

\(\displaystyle+{T}_{{1}}{r}_{{1}}-{T}_{{2}}{r}_{{2}}={\left({\frac{{{1}}}{{{2}}}}\right)}{\left({m}_{{2}}{{r}_{{2}}^{{2}}}\right)}{\left({\frac{{{a}}}{{{r}_{{2}}}}}\right)}\)

apply newtons second law oftranslational motion to the crate then we get

\(\displaystyle+{T}_{{2}}-{m}_{{3}}{g}={m}_{{3}}{a}\)

now solving for \(\displaystyle{T}_{{1}}\) from equation(1) and substituting the result in (2) then solving (2) for \(\displaystyle{T}_{{2}}\) and substituting the result in (3) we get the following value of the torque as

\(\displaystyle{T}={r}_{{1}}{\left[{a}{\left({m}_{{1}}+{\left({\frac{{{1}}}{{{2}}}}{m}_{{2}}+{m}_{{3}}\right)}+{m}_{{3}}{g}\right]}\right.}\)

= (0.76 m) {(1.5 m /s2) [182 kg + (1 / 2) (130 kg) + (150 kg)] + (150 kg)(9.80 \(\displaystyle\frac{{m}}{{s}^{{2}}}\))

= ....... N.m

\(\displaystyle{T}_{{1}}\) as magnitude of tension in thecord between the drum and the pulley

the net torque is calculated accordingto the equation

\(\displaystyle\sum{T}={I}\alpha\)

so the net torque exerted on the drum will be

\(\displaystyle\sum{T}={I}_{{1}}\alpha_{{1}}\)

Where \(\displaystyle{I}_{{1}}\) is the moment of inertia of the drum

\(\displaystyle\alpha_{{1}}\) is the angular acceleration of the drum

if we take that the cable does not slip then we can write

\(\displaystyle-{T}_{{1}}{r}_{{1}}+{T}={\left({m}_{{1}}{{r}_{{1}}^{{2}}}\right)}{\left({\frac{{{a}}}{{{r}_{{1}}}}}\right)}\)

in the above equation

T is the counter clockwise torque providedby the motor and a is the acceleration of the cord as from the given we can see that

\(\displaystyle{a}={1.3}\frac{{m}}{{s}^{{2}}}\)

in the next step we take \(\displaystyle{T}_{{2}}\) as magnitude of tension in the cordbetween the crate and the pulley and \(\displaystyle{I}_{{2}}\) as the moment of inertia of the pulley next we apply newtons second law of motion to thepulley then we get

\(\displaystyle+{T}_{{1}}{r}_{{1}}-{T}_{{2}}{r}_{{2}}={\left({\frac{{{1}}}{{{2}}}}\right)}{\left({m}_{{2}}{{r}_{{2}}^{{2}}}\right)}{\left({\frac{{{a}}}{{{r}_{{2}}}}}\right)}\)

apply newtons second law oftranslational motion to the crate then we get

\(\displaystyle+{T}_{{2}}-{m}_{{3}}{g}={m}_{{3}}{a}\)

now solving for \(\displaystyle{T}_{{1}}\) from equation(1) and substituting the result in (2) then solving (2) for \(\displaystyle{T}_{{2}}\) and substituting the result in (3) we get the following value of the torque as

\(\displaystyle{T}={r}_{{1}}{\left[{a}{\left({m}_{{1}}+{\left({\frac{{{1}}}{{{2}}}}{m}_{{2}}+{m}_{{3}}\right)}+{m}_{{3}}{g}\right]}\right.}\)

= (0.76 m) {(1.5 m /s2) [182 kg + (1 / 2) (130 kg) + (150 kg)] + (150 kg)(9.80 \(\displaystyle\frac{{m}}{{s}^{{2}}}\))

= ....... N.m