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 1.70 kg and diameter 0.400 m .

Part A After the system is released, find the horizontal tension in the wire. |Th| = N

Part B After the system is released, find the vertical tension in the wire. |Tv| = N

Part C After the system is released, find the acceleration of the box. a = m/s2

Part 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. Fx,Fy = N

Explanation / Answer

Let’s determine the net horizontal force on the box and the net vertical force on the other object. For the box, the net force is the tension. For the other object, the net force is equal to its weight minus the tension.

For the box, T1 = 12 * a
For the other object, 5 * 9.8 – T2 = 5 * a
T2 = 49 – 5 * a

The two tension forces are producing the torque that is causing the pulley to accelerate.

Torque = I *
I = ½ * 1.7 * 0.26^2 = 0.05746
Since we need to determine the acceleration of the box, let’s convert to a by multiplying by the radius.

Torque = 0.05746 * 0.26 * a = 0.0149396

Torque = tension * radius
For T2, torque = T2 * 0.26 For T1, torque = T1 * 0.26
Net torque = T2 * 0.23 – T1 * 0.23 = (T2 – T1) * 0.26
Set this equal to 0.0149396.

(T2 – T1) * 0.26 = 0.0149396
T2 – T1 = 0.05746
T2 = T1 + 0.05746

Let’s substitute T1 + 0.05746 for T2 in the following equations


T1 + 0.05746 = 49 – 5 * a
Subtract 0.05746 from both sides.
T1 = 48.94254 – 5 * a
T1 = 12 * a
Set these two equations equal to each other and solve for a.
12 * a = 48.94254 – 5 * a
17 * a = 48.94254
a = 48.94254 ÷ 17 = 2.878972941 m/s^2
Let’s use the acceleration to determine the value of each tension force.

T1 = 12 * 2.878972941 = 34.54767529 N
T2 = 49 – 5 * 2.878972941 = 34.60513529 N
T2 – T1 = 34.60513529 – 34.54767529 = 0.05745939
Since this is almost 0.05746, I believe the tension forces are correct!

Since T1 is the only horizontal force on the pulley, the magnitude of the horizontal on the axle is equal to T1. The axle has two vertical forces, its weight and T2. The total vertical force is the sum of these two forces.

Weight = 1.7 * 9.8 = 16.66 N
Total vertical force = 34.60513529 + 16.66

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