The crane shown in the drawing is lifting a 182-kg crate upward with an accelera

Question

The crane shown in the drawing is lifting a 182-kg crate upward with an acceleration of1.5 m/s2. The cable from thecrate passes over a solid cylindrical pulley at the top of theboom. 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. Theengine applies a counterclockwise torque to the drum in order towind up the cable. What is the magnitude of this torque? Ignore themass of the cable.
1 N ·m

Explanation / Answer

   let us represent    T1 as magnitude of tension in thecord between the drum and the pulley    the net torque is calculated accordingto the equation     = I    so the net torque exerted on the drum willbe     = I11    where I1 is the moment of inertia ofthe drum    1 is the angular acceleration ofthe drum    if we take that the cable does not slip thenwe can write    - T1 r1 + =(m1 r12) (a / r1)............ (1)    in the above equation    is the counter clockwise torque providedby the motor and a is the acceleration of the cord    as from the given we can seethat        a = 1.3 m / s2    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 ofthe pulley    next we apply newtons second law of motion to thepulley then we get    + T1 r1 -T2 r2 = (1 / 2) (m2r22) (a / r2) ...........(2)    apply newtons second law oftranslational motion to the crate then we get    + T2 - m3 g = m3a ............ (3)    now solving for T1 from equation(1) and substituting the result in (2) then solving (2) forT2 and    substituting the result in (3) we get thefollowing value of the torque as    = r1 [a (m1 + (1 /2) m2 + m3) + m3 g]       = (0.76 m) {(1.5 m /s2) [182 kg + (1 / 2) (130 kg) + (150 kg)] + (150 kg)(9.80 m / s2)       = ....... N . m    where I1 is the moment of inertia ofthe drum    1 is the angular acceleration ofthe drum    if we take that the cable does not slip thenwe can write    - T1 r1 + =(m1 r12) (a / r1)............ (1)    in the above equation    is the counter clockwise torque providedby the motor and a is the acceleration of the cord    as from the given we can seethat        a = 1.3 m / s2    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 ofthe pulley    next we apply newtons second law of motion to thepulley then we get    + T1 r1 -T2 r2 = (1 / 2) (m2r22) (a / r2) ...........(2)    apply newtons second law oftranslational motion to the crate then we get    + T2 - m3 g = m3a ............ (3)    now solving for T1 from equation(1) and substituting the result in (2) then solving (2) forT2 and    substituting the result in (3) we get thefollowing value of the torque as    = r1 [a (m1 + (1 /2) m2 + m3) + m3 g]       = (0.76 m) {(1.5 m /s2) [182 kg + (1 / 2) (130 kg) + (150 kg)] + (150 kg)(9.80 m / s2)       = ....... N . m

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