A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.0

Question

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 without slippage over a frictionless pulley (the figure (Figure 1) ). The pulley has the shape of a uniform solid disk of mass 1.60 kg and diameter 0.580 m

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.

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.

Express your answers separated by a comma.

Explanation / Answer

m1 = 12.0 Kg
m2 = 5.0 Kg

Let the Horizontal Tension be, T1

T1 = m1*a
T1 = 12.0*a -----1

Let the Vertical Tension be, T2
5.0*g - T2 = 5.0*a -------2

For Pulley,
(T2 - T1)*r = I*(a/r)
I is Moment of Inertia = 1/2*M*r^2
(5.0*g - 5.0*a - 12.0*a) = 1/2*1.6*r^2 * (a/r^2)
(5.0*g -17.0*a) =1/2*1.6 * a
a = 2.75 m/s^2

(a)
T1 = 12.0*2.75 N
T1 = 33 N
Horizontal Tension, T1 = 33 N

(b)
T2 = 5.0*9.8 - 5.0*2.75
T2 = 35.3 N
Vertical Tension, T2 = 35.3 N

(c)
Acceleration of the box, a = 2.75 m/s^2

(d)
Magnitude of the horizontal component of the force F that the axle exerts on the pulley, Th = 33 N
Magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of Pulley = 35.3 + 1.6*9.8
= 50.98 N

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