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 cylindr

cistG 2021-04-21 Answered
The crane shown in the drawing is lifting a 182-kg crate upward with an acceleration of 1.5ms2. 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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Expert Answer

Derrick
Answered 2021-04-23 Author has 94 answers

Let us represent
T1 as magnitude of tension in thecord between the drum and the pulley
the net torque is calculated accordingto the equation
T=Iα
so the net torque exerted on the drum will be
T=I1α1
Where I1 is the moment of inertia of the drum
α1 is the angular acceleration of the drum
if we take that the cable does not slip then we can write
T1r1+T=(m1r12)(ar1)
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
a=1.3ms2
in the next step we take T2 as magnitude of tension in the cordbetween the crate and the pulley and I2 as the moment of inertia of the pulley next we apply newtons second law of motion to thepulley then we get
+T1r1T2r2=(12)(m2r22)(ar2)
apply newtons second law oftranslational motion to the crate then we get
+T2m3g=m3a
now solving for T1 from equation(1) and substituting the result in (2) then solving (2) for T2 and substituting the result in (3) we get the following value of the torque as
T=r1[a(m1+(12m2+m3)+m3g]
(0.76m)(1.5ms2)[182kg+(12)(130kg)+(150kg)]+(150kg)(9.80ms2)]
= ....... N.m

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