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# A 12.0-kg box resting on a horizontal, frictionless ## Problem 17E Chapter 10

University Physics | 13th Edition

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Problem 17E

A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes over a frictionless pulley (?Fig. E10.16?). The pulley has the shape of a uniform solid disk of mass 2.00 kg and diameter 0.500 m. After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

Step-by-Step Solution:

Solution 17E Step 1: Given data: Mass of the pulley m = 2.0 kg Weight of the pulley w = mg = (2.0 kg)(9.8 m/s ) = 18 N Radius r = d/2 = 0.500 m/2 = 0.25 m Mass of the box M = 12.0 kg Mass of attached weight m = 5.0 kg a 2 2 2 Moment of inertia I = 1/2 mr = 1/2(2.0 kg)(0.25 m) = 0.0625 kg/m Acceleration of box=a Step 2: Net force on box F = ma Tension in the horizontal portion of wire T =h12 a *(1) Tension in the vertical portion of wire = T v Weight suspended is w = m g =a(5.00 kg)(9.8 m/s ) = 49 N But net force suspended on weight F = 5a 49 T = 5a (2) v If is angular acceleration of the pulley = linear acceleration/radius = a/0.25 m Net torque = [T vT ] rh=*[T T ]v.25 h Net torque = [49 5a 12 a]0.25 m Net torque = [49 17a]0.25 m Net torque = I = Ia/r = 0.0625a/0.25 = 0.25a 0.25 a = [49 17a]0.25 (3) 2 a = 2.578 m/s Step 3: (a). orizontal tension is given by Th= 12 a* T h 12 (2*578 m/s ) 2 T = 30.94 N h vertical tension is given by T = 49 5a v 2 T v 49 5(2.578 m/s ) T v 36.11 N

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