Force Systems accelerate together Combination Systems connected masses

Force Systems accelerate together • Combination Systems – connected masses • Horizontal Pulley • Atwood’s Machine

Connected Masses What forces can you identify acting each box? What is the net force on the system?

A constant net force, F, accelerates the entire system’s mass. Fnet = mtota for the system a = Fnet. m 1 + m 2 + m 3.

Each box has separate forces acting on it. Sketch free body diagrams for each mass ignore vertical forces. Assign 1 direction as positive (right). Write the Fnet equation for each, find acceleration. Isolate each masses to find T 1 & T 2. m 1 a = T 1. m 2 a = T 2 - T 1. m 3 a = F – T 2.

Ex 1: Connected Masses: Given a Fnet of 20 N and masses of 4, 3, and 1 kg, find the acceleration of the system and the tension in each cord. a = Fnet. m 1 + m 2 + m 3

Find system acceleration: a = Fnet m 1 + m 2 + m 3. 20 N (4 + 3 + 1) kg a = 2. 5 m/s 2.

Use the free body diagram & known acceleration to find the tension in each cord. 4 kg T 1 = m 1 a = (4 kg)(2. 5 m/s 2) = 10 N. F -T 2 = m 3 a or F - m 3 a = T 2. 20 N - (1 kg)(2. 5 m/s 2) = 1 kg 17. 5 N

Check the calculation using the 3 rd mass. T 2 – T 1 = m 2 a 17. 5 N – 10 N = 7. 5 N m 2 a = (3 kg)(2. 5 m/s 2) = 7. 5 N. It is correct!!

m 1 a = T 1. m 2 a = T 2 - T 1. • T 1 = 10 N • T 2 = 17. 5 N

Horizontal Pulley.

The masses accl together, the tension is uniform, accl direction is positive.

Sketch free body diagrams for each mass separately. Write Newton’s 2 nd Law equation for each. -T. F n. M 1. m 1 g m 1 a = T +T. M 2. m 2 g m 2 a = m 2 g - T

Add the equations: m 1 a + m 2 a = T + m 2 g – T T cancels. m 1 a + m 2 a = m 2 g Factor a & solve

a = m 2 g m 1 + m 2 Solve for a, and use the acceleration to solve for the tension pulling one of the masses. m 1 a = T

Ex 2: Horizontal Pulley: Given a mass of 4 kg on a horizontal frictionless surface attached to a mass of 3 kg hanging vertically, calculate the acceleration, and the tension in the cord. Compare the tension to the weight of the hanging mass, are they the same?

a = 4. 2 m/s 2 T = 16. 8 N mg = 30 N, it is less than the tension.

Ex: Given horizontal pulley system where m 1 = 2 kg, m 2 = 5 kg, and m, the coefficient of friction between m 1 and the counter is 0. 35, sketch the free body diagrams for m 1 and m 2, calculate the acceleration of the system, and find T.

To practice problems go to: Hyperphysics site. Click Mechanics, Newton’s Laws, Standard problems, then the appropriate symbol. http: //hyperphysics. phyastr. gsu. edu/hbase/hph. html#mechcon

Atwood’s Machine Use wksht.

Given Atwood’s machine, m 1 = 2 kg, m 2 = 4 kg. Find the acceleration and tension.

Sketch the free body diagram for each.

Boxes in Contact

Since F is the only force acting on the two masses, it determines the acceleration of both: The force F 2 acting on the smaller mass may now be determined.

Using the previously determined accl, the force F 2 acting on the smaller mass is F 2 = m 2 a By Newton’s 3 rd Law, F 2 acts backward on m 1. The net force, F 1, on m 1 is: F m 1 F 2

Given a force of 10 N applied to 2 masses, m 1 =5 kg and m 2 =3 kg, find the accl and find F 2 (the contact force) between the boxes. a = 1. 25 m/s 2 F 2 = m 2 a F 2 = 3. 75 N

Given a force of 100 N on 100 1 kg boxes, what is the force between the 60 th and 61 st box. 100 -N 1 -kg

Find a for system. a = 1 m/s 2. F 2 must push the remaining 40 boxes or 40 kg. F = 40 kg (1 m/s 2. ) 40 N.

Ignoring friction, derive an equation to solve for a and T for this system: Begin by sketching the free body diagram Write the equations for each box Add them. Solve for accl