The crane shown in the drawing is lifting a 182-kg crate upward with an acceleration of 1.5m/s^2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound ontoa hollow cylindrical drum that is mounted on the deck of the crane.The mass of the drum is 150 kg, and its radius is 0.76 m. The engine applies a counter clockwise torque to the drum in order towind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

The crane shown in the drawing is lifting a 182-kg crate upward with an acceleration of 1.5m/s^2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound ontoa hollow cylindrical drum that is mounted on the deck of the crane.The mass of the drum is 150 kg, and its radius is 0.76 m. The engine applies a counter clockwise torque to the drum in order towind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

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asked 2021-04-21
The crane shown in the drawing is lifting a 182-kg crate upward with an acceleration of \(\displaystyle{1.5}\frac{{m}}{{s}^{{2}}}\). The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound ontoa hollow cylindrical drum that is mounted on the deck of the crane.The mass of the drum is 150 kg, and its radius is 0.76 m. The engine applies a counter clockwise torque to the drum in order towind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

Answers (1)

2021-04-23
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
0

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