Question

Determine the power output of the draw-works motor M necessary to lift the 600-lb drill pipe upward with a constant speed of 4 ft/s. The cable is tied to the top of the oil rig, wraps around the lower pulley, then around the top pulley, and then to the motor.

Answer 1

`\therefore` The power output of the draw-works motor is 4.36 hp

Step-by-step explanation:

Given that,

• 600 lb of weight of the drill pipe
• 4 ft/s constant velocity of the motor
• M power output of the motor

As the problem given, the cable is tied of the top on the oil rig and it wraps around the lower pulley and the top pulley. So, we will be using the formula below:

• 
  `\boxed{\rm{2s_p + s_M = l}}`

Differentiate it but with respect to the time. Therefore,

• 
  `2v_p +v_M = 0`

Thus,

• 
  `V_M = -2v_p = -2(-4) = 8 \ ft/s`

Thus, the power output should be

`P = Fv = \bigg((600)/(2)\bigg) \cdot 8 = 2400 \ lb \cdot ft/s = (2400)/(550) = 4.36 \ hp`

`\therefore` The power output of the draw-works motor is 4.36 hp