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Viefleur [7K]
3 years ago
13

The grooved pulley of mass m is acted on by a constant force F through a cable which is wrapped securely around the exterior of

the pulley. The pulley supports a cylinder of mass M which is attached to the end of a cable which is wrapped securely around an inner hub. If the system is stationary when the force F is first applied, determine the upward velocity v of the supported mass after 4 seconds. Use the values m = 30 kg, M = 11 kg, ro = 345 mm, ri = 230 mm, kO = 250 mm, and F = 85 N. Assume no mechanical interference for the indicated time frame and neglect friction in the bearing at O. What is the time-averaged value of the force T in the cable which supports the 11-kg mass?
Physics
1 answer:
Sidana [21]3 years ago
3 0

Answer:

Answer; v= 1.2654m/s

T= 110.76N

Explanation:

Apply Momentum Principle

Fdtro - Mgridt = Iow +Mvr

Fdtro - Mgridt = mK2 v/r1 + Mvr1

85 x 3x 0.345 -11 x 9.81 x 0.23 x 3 =30 x 0.25 x 0.25 x v/0.23 + 11 x v x 0.23 =

v = 1.2654m/s

To find the timed average value

Tdt -Mgdt =MV

T x 3 - 11 x 9.81 x 3 = 11 x 0.778

T= 110.76N

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x(2.50 s)=(17.5 cm) \cos ( (52.7 rad/s)(2.50 s))=17.2 cm
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A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo
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Explanation:

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    Mass of big bucket (M) = 12 kg

    Initial velocity (v_{o}) = 0 m/s

    Final velocity (v_{f}) = ?

  Height H_{o} = h_{f} = 2 m

and,    H_{f} = h_{o} = 0 m

Now, according to the law of conservation of energy

         starting conditions = final conditions

  \frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}

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Answer:

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3 TA - 3 T = T - TB

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T = 1.75 TB

8 0
2 years ago
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