Two masses are suspended by a cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m₁, has a mass of 7.0 kg and the other mass, m₂, has a mass of 3.0 kg. The pulley turns on a shaft through the center of the pulley, which supports the pulley and all the masses. The vertical force of the shaft on the pulley is A) 49 N upward. B) 49 N downward. C) 82 N upward. D) 82 N downward. E) 98 N upward.

Question

Quizplus · Accepted Answer

The acceleration of the system can be determined by analyzing the forces acting on each object. The force of gravity on object 1 is m1g = (7.0 kg)(9.8 m/s^2) = 68.6 N and the force of gravity on object 2 is m2g = (3.0 kg)(9.8 m/s^2) = 29.4 N. Since the cord is taut, the tension in the cord on both sides of the pulley is equal in magnitude. Let T be the tension in the cord. The force balance on object 1 gives:

T - m1g = m1a

The force balance on object 2 gives:

m2g - T = m2a

Since the pulley is massless and the cord is massless, the tension in the cord is the same on both sides of the pulley. Thus, the net force on the pulley is equal to the tension in the cord, and the direction of the net force is the direction of the tension in the cord. The force balance on the pulley gives:

2T = Iα

where I is the moment of inertia of the pulley and α is its angular acceleration. Since the pulley rotates about its center and all the masses are perpendicular to the axis of rotation, the moment of inertia of the pulley is given by I = (1/2)mr^2, where r is the radius of the pulley. Thus,

2T = (1/2)mr^2α

The relationship between the linear acceleration of the masses and the angular acceleration of the pulley can be written as a = rα, where a is the linear acceleration of the masses. Substituting this into the previous equation yields:

2T = (1/2)m(ar) = mar

Substituting the expressions for the tensions in the cord from the force balances on the masses yields:

(m1g - T) + T = m1a

T - (m2g - T) = m2a

Solving for T in each equation and setting them equal to each other yields:

m1g - m1a = m2a + m2g

Solving for a gives:

a = (m1 - m2)g / (m1 + m2)

Substituting the given values yields:

a = (7.0 kg - 3.0 kg)(9.8 m/s^2) / (7.0 kg + 3.0 kg) = 3.92 m/s^2

The net force on the pulley is then:

Fnet = 2T = (1/2)mr(a / r) = ma = (10.0 kg)(3.92 m/s^2) = 39.2 N

Since the force of gravity on the pulley is 98 N and the net force on the pulley is upward, the vertical force of the shaft on the pulley must be:

Fn = Fgrav + Fnet = 98 N + 39.2 N = 137.2 N

The force of the shaft on the pulley is upward, so the answer is (C) 82 N upward.

Two Masses Are Suspended by a Cord That Passes Over