A string that passes over a pulley has a 0.373 kg mass attached to one end and a 0.620 kg mass attached to the other end. The pulley, which is a disk of radius 9.50 cm, has friction in its axle. What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium?

To prevent the pulley from turning, the frictional torque must equal the difference of the weight-applied torques.

Tf = (0.620 - 0.373)kg*9.8 m/s^2*0.095 m
= 0.230 kg*m^2/s^2 (Newton-meter)

Well, since the system is in static equilibrium, we can say that the sum of the torques acting on the pulley must be equal to zero. Now, let me think about a joke related to pulleys and friction... Ah! Here it is:

Why did the pulley break up with the friction?

Because it found someone more supportive!

Now, back to your question. To determine the magnitude of the frictional torque, we need to consider the forces acting on the pulley. We have the weight of the masses and the frictional torque.

However, without knowing the coefficient of friction, it is not possible to calculate the exact value of the frictional torque. So, I'm afraid I can't give you the numerical answer.

But hey, at least you got a joke out of it!

To find the magnitude of the frictional torque exerted by the axle in static equilibrium, we need to analyze the forces acting on the system.

First, let's consider the forces acting on the 0.373 kg mass.
- The weight (W) of the mass is given by W = m * g, where m is the mass and g is the acceleration due to gravity.
- The tension (T1) in the string attached to this mass will be equal to W since there is no acceleration in the vertical direction.

Next, let's consider the forces acting on the 0.620 kg mass.
- The weight (W) of the mass is given by W = m * g.
- The tension (T2) in the string attached to this mass will be equal to W since there is no acceleration in the vertical direction.

Since the system is in static equilibrium, the net torque about the axle must be zero. The torque (τ) exerted by a force (F) about an axis of rotation at a distance (r) from the axis is given by τ = F * r * sin(θ), where θ is the angle between the force and the radial direction.

In this case, the torque exerted by the tension forces T1 and T2 will be countered by the frictional torque (τ_friction) exerted by the axle.

Now, let's calculate the torques exerted by the tension forces.
- The torque exerted by T1 is given by τ1 = T1 * r, where r is the radius of the pulley.
- The torque exerted by T2 is given by τ2 = T2 * r.

Since T1 = W and T2 = W, we can rewrite the torques as follows:
- τ1 = W * r
- τ2 = W * r

To maintain static equilibrium, the frictional torque exerted by the axle must be equal in magnitude and opposite in direction to the sum of the torques exerted by the tension forces. Therefore:
τ_friction = τ1 + τ2

Substituting the torques we calculated earlier, we get:
τ_friction = (W * r) + (W * r)
τ_friction = 2W * r

Now, let's substitute the values given in the problem:
- Mass of 0.373 kg: m1 = 0.373 kg
- Mass of 0.620 kg: m2 = 0.620 kg
- Radius of the pulley: r = 9.50 cm = 0.0950 m
- Acceleration due to gravity: g = 9.8 m/s^2

Calculating the weights:
- Weight of m1: W1 = m1 * g
- Weight of m2: W2 = m2 * g

Substituting the values and calculating the frictional torque:
τ_friction = 2(W1 * r) + 2(W2 * r)
τ_friction = 2[(m1 * g) * r] + 2[(m2 * g) * r]

Calculating the final result:
τ_friction = 2[(0.373 kg * 9.8 m/s^2) * 0.0950 m] + 2[(0.620 kg * 9.8 m/s^2) * 0.0950 m]

To find the magnitude of the frictional torque exerted by the axle, we need to understand the conditions for static equilibrium and the torque equation.

Conditions for static equilibrium:

1. The sum of all forces acting on the system must be zero.
2. The sum of all torques acting on the system must be zero.

Torque equation:

Torque (τ) = Force (F) * Perpendicular Distance (r)

To find the frictional torque exerted by the axle, we need to analyze the forces and distances involved in this system.

1. Forces involved:
- Tension force from the string (Tension in side with 0.373 kg mass) = m1 * g (where m1 is the mass of the 0.373 kg mass and g is the acceleration due to gravity)
- Tension force from the string (Tension in side with 0.620 kg mass) = m2 * g (where m2 is the mass of the 0.620 kg mass)

2. Distances involved:
- Perpendicular distance between the point of rotation (center of the pulley) and the tension force from the string on the side with 0.373 kg mass = radius of pulley (r)

Since the pulley is in rotational equilibrium, the torque from the tension in side with 0.373 kg mass must balance out the torque from the tension in side with 0.620 kg mass.

Setting up the equation:

τ1 (from side with 0.373 kg mass) - τ2 (from side with 0.620 kg mass) = 0

Using the torque equation, we can express this equation as:

(Tension in side with 0.373 kg mass) * r - (Tension in side with 0.620 kg mass) * r = 0

Now, substitute the values into the equation:

(m1 * g) * r - (m2 * g) * r = 0

Simplifying further:

(m1 - m2) * g * r = 0.373 kg * 9.8 m/s^2 * 0.095 m - 0.620 kg * 9.8 m/s^2 * 0.095 m

Now, we can calculate the magnitude of the frictional torque by multiplying the above result by -1, as the torque from the axle must oppose the tension:

Frictional Torque = -((m1 - m2) * g * r)

Plugging in the values:

Frictional Torque = -((0.373 kg - 0.620 kg) * 9.8 m/s^2 * 0.095 m)

Calculate and simplify:

Frictional Torque = -(0.620 kg * 9.8 m/s^2 * 0.095 m - 0.373 kg * 9.8 m/s^2 * 0.095 m)