A 12.0- box resting on a horizontal, frictionless surface is attached to a 5.00-

Question

A 12.0- box resting on a horizontal, frictionless surface is attached to a 5.00- weight by a thin, light wire that passes without slippage over a frictionless pulley. The pulley has the shape of a uniform solid disk of mass 1.50 and diameter 0.420. (a). After the system is released, find the horizontal tension in the wire. (b) After the system is released, find the vertical tension in the wire. (c) After the system is released, find the acceleration of the box. (d) After the system is released, find magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

Explanation / Answer

Start with FBD's of the hanging mass, pulley and sliding mass Hanging: 5*9.81-T1=5*a where T1 is the tension in the wire above the hanging mass rearrange a bit T1=5*(9.81-a) The pulley (note that the axle does not introduce a torque on the pulley) 0.540*(T1-T2)= (1.50*0.420^2/2)*(a/0.420) note also that a=a/r simplify a bit T1-T2=0.75*a The sliding mass T2=12.0*a let's collect our three equations and solve for our three unknowns: T1, T2, and a T1=5*(9.81-a) T1-T2=0.75*a T2=12.0*a Combining the first and last into the middle: 5*9.81/17.75=a a=2.76 m/s T1=35.23 N T2=33.16 N Okay, now lets look at the vertical and horizontal forces on the axle of the pulley Vertical: T1+1.5*9.81 49.95 N So the axle reacts with an upward force of 49.95 N Horizontal T2 33.16 N The axle reacts with a force to the right of 33.16 N

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