The cable passes over a solid cylindrical pulley at the top ofthe boom. The pull

Question

The cable passes over a solid cylindrical pulley at the top ofthe boom. The pulley has a mass of 130 kg. The cable is then woundaround a hollow cylindrical drum mounted on the deck of the crane.The mass of the drum is 150 kg and its radius is .76 m. The engineapplies a counterclockwise torque to the drum in order to wind upthe cable. What is the magnitude of the torque?



Work: A: Pulley Sums of Torques= I(alpha) = mr^2(a/r)= 1/2(130)r^2(1.2/r)=65r(1.2)=.................78r Sums of Forces= ma=180(9.8)-180(1.2)=........................................1548

B: Drum Sums of Torque= I(alpha) = 87(alpha)= 87(a/r)=.................................................114a Sums of Forces= ma = 150(a)
Do we add the two two forces and the two torques together? ordo we have too many variables?

Explanation / Answer

   let us represent
   T1 as magnitude of tension in the cordbetween the drum and the pulley
   the net torque is calculated according tothe equation
    = I
   so the net torque exerted on the drum will be
    = I11
   where I1 is the moment of inertia of thedrum
   1 is the angular acceleration of thedrum
   if we take that the cable does not slip then wecan write
   - T1 r1 + =(m1 r12) (a / r1)............ (1)
   in the above equation
   is the counter clockwise torque provided by themotor and a is the acceleration of the cord
   as from the given we can seethat   
   a = 1.2 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 of translational motionto the crate then we get
   + T2 - m3 g = m3 a............ (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) {(1.2 m /s2) [150 kg + (1 / 2) (130 kg) + (180 kg)] + (180 kg)(9.80 m / s2)
      = ....... N . m

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