Consider the system shown in the diagram. The pulley is a uniform cylinder with mass m3 = 0.60 kg and radius R = 4.0 cm, the other two masses are m1 = 2.5 kg and m2 = 1.0 kg, and á = 30o. Assume the rope is massless, there is no slipping of the rope on the pulley, and there is no friction between m1 and the incline.
(a) What is the acceleration of m1 and m2 (both magnitude and direction)? What is the angular acceleration of the pulley? What are the tensions in the rope (both to the left of the pulley and to the right of the pulley?
If what I'm thinking it looks like is true a force of 12.25 N is pulling the 9.8 N up for an acc of 0.7. That net 2.45 force causes a torque of .098 on the pulley. I think (better look it up) I = 1/2mr^2 for a solid cylinder, so alpha = tau over I. Do an FBD on the hanging box for tension.
Bob? Damon? am I talking out my butt here?
To find the acceleration of m1 and m2, as well as the angular acceleration of the pulley, and the tensions in the rope, we will use Newton's laws of motion and torque.
1. Acceleration of m1 and m2:
First, let's consider the forces acting on m1 and m2.
- There is the gravitational force acting vertically downwards on both m1 and m2, given by mg, where g is the acceleration due to gravity.
- There is a tension force T1 acting horizontally on m1, and a tension force T2 acting horizontally on m2.
For m1:
- The vertical component of the gravitational force does not contribute to the horizontal motion, so we only consider the horizontal forces.
- The net force on m1 is the difference between the tension force T1 (which points to the left) and the horizontal component of the gravitational force (which points to the right).
- The net force on m1, F1, is given by F1 = T1 - m1 * g * sin(α), where α is the angle of the incline.
For m2:
- The net force on m2 is the difference between the horizontal component of the gravitational force (which points to the left) and the tension force T2 (which points to the right).
- The net force on m2, F2, is given by F2 = m2 * g * sin(α) - T2.
Since the masses are connected by a rope, their accelerations will be the same, and we can equate the net forces to the mass times acceleration (F = m * a).
2. Angular acceleration of the pulley:
The torque on the pulley is the difference between the tension forces acting on opposite sides of the pulley. The torque, τ, is given by τ = T1 * R - T2 * R, where R is the radius of the pulley.
The moment of inertia of a uniform cylinder rotating about its center is given by I = (1/2) * m * R^2.
The torque is related to the angular acceleration, α, by τ = I * α.
So, we have T1 * R - T2 * R = (1/2) * m3 * R^2 * α.
3. Tensions in the rope:
To find the tensions in the rope, we can use the equations we derived for m1 and m2:
- From the equation for m1: T1 = m1 * a + m1 * g * sin(α).
- From the equation for m2: T2 = m2 * g * sin(α) - m2 * a.
Note that the tension T1 is to the left of the pulley, while T2 is to the right of the pulley.
By solving these equations simultaneously, we can find the values of acceleration of m1 and m2, angular acceleration of the pulley, and the tensions in the rope.