a 200 kg block is lifted at a constant speed of 0.5 m/s by a steel cable that passes over a massless, frictionless pulley to a motor driven

Question

a 200 kg block is lifted at a constant speed of 0.5 m/s by a steel cable that passes over a massless, frictionless pulley to a motor driven wheel. the radius of the wheel is 30cm

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Khánh Gia 3 years 2021-08-30T02:08:41+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-30T02:10:38+00:00

    Questions

    (a) What is the magnitude of the force exerted by the cable?

    (b) What is the magnitude of the torque exerted by the cable on the winch drum?

    (c) What is the angular velocity of the winch drum?

    (d) What power must be developed by the motor to drive the winch drum?

    Answer:

    (a) 1962 N

    (b) 588.6 N.m

    (c) 1.67 rads/s

    (d) 982.962 W

    Explanation:

    (a)

    Force, F=mg where m is the mass of the block and g is acceleration due to gravity. Taking the value of acceleration due to gravity as 9.81 m/s2 and substituting 200 Kg for mass then the force exerted on the cable will be

    F=200*9.81=1962 N

    (b)

    Torque is a product of force and distance. Since the force is already derived in part (a) above and the radius is given as 30 cm converted to m we have 0.3 m then torque, T= 1962*0.3=588.6 N.m

    (c)

    Angular velocity= \frac {v}{r} where v is the speed of block and r is the radius of the wheel. Substituting 0.5 m/s for v and 0.3 m for r then the angular velocity will be \frac {0.5}{0.3}=1.6667\approx 1.67 rads/s

    (d)

    Power, P is a product of torque and angular velocity and since torque is already given in part b above as 588.6 N and the angular velocity in part c then power, P= 588.6*1.67=982.962 W

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