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

For any force system you must sum forces. Fnet = SF = F1 + F2 … For any force system you must sum forces. Fnet = SF = F1 + F2 … ma = F1 + F2 …

Connected Masses What forces can you identify acting on the boxes? Let’s call F the net force on the system

Fnet must accelerate the entire system’s mass together. Fnet = mtota

Given the masses and Fnet: sketch free body diagrams for each mass ignore vertical forces. Assign 1 direction as positive (right). Write the Fnet equation for each, find acceleration. then isolate each masses to find T1 & T2. m1a = T1. m2a = T2 - T1. m3a = F – T2.

Add the equations. Factor & make cancellations. m1a = T1. m2a = T2 - T1. m3a = F – T2. m1a + m2a + m3a = T1 + T2 – T1 + F – T2. a (m1 + m2 + m3) = F.

Equation to solve problems in connected masses. a = Fnet. m1 + m2 + m3

Ex 1: Connected Masses: Given a Fnet of 20N and masses of 4, 3, and 1 kg, find the acceleration of the system and the tension in each cord. 20 N (4 + 3 + 1) kg a = 2.5 m/s2.

Use the free body diagram & known acceleration to find the tension in each cord. 4 kg m1a = T1 = (4 kg)(2.5 m/s2) = 10 N. m3a = F -T2 Fnet - m3a = T2. 20N - (1 kg)(2.5 m/s2) = 17.5 N 1 kg

Check the calculation using the 3rd mass. T2 – T1 = m2a. 17 Check the calculation using the 3rd mass. T2 – T1 = m2a 17.5 N – 10 N = 7.5 N m2a = (3 kg)(2.5 m/s2) = 7.5 N. It is correct!!

m1a = T1. m2a = T2 - T1. T1 = 10 N T2 = 17.5 N

Horizontal Pulley.

The masses accl together, the tension is uniform, accl direction is positive. Sketch the free body diagram each mass.

Consider any F contributing to acceleration + Consider any F contributing to acceleration +. Any F opposing acceleration is - . Apply Newton’s 2nd Law equation for each. M2. -T. m2g M1. +T. Fn. m1g m1a = T m2a = m2g - T

Add the equations: m1a + m2a = T + m2g – T T cancels. m1a + m2a = m2g Factor a & solve

a = m2g m1 + m2 Solve for a, and use the acceleration to solve for the tension pulling one of the masses. m1a = 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 = m2g m1 + m2 30 N 7 kg a = 4.3 m/s2

m1a = T (4 kg)(4.3 m/s2) T = 17 N mg = 30 N, tension less than weight.

Atwood’s Machine Use wksht.

Sketch the free body

The equation m1a = T – m1g m2a = m2g – T a (m1 +m2) = m2g - m1g a = m2g - m1g m tot

Given Atwood’s machine, m1 = 2 kg, m2 = 4 kg Given Atwood’s machine, m1 = 2 kg, m2 = 4 kg. Find the acceleration and tension. a = 3.3 m/s2. T = 26 N

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 F2 acting on the smaller mass may now be determined.

Using the previously determined accl, the force F2 acting on the smaller mass is F2 = m2a

By Newton’s 3rd Law, F2 acts backward on m1. The force on m1 is: The net force, F1, on m1 is: m1 F2 F

Given a force of 10N applied to 2 masses, m1 =5 kg and m2 =3kg, find the accl and find F2 (the contact force) between the boxes. a = 1.25m/s2 F2 = 3.75 N

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

Find a for system. F2 must push the remaining 40 boxes or 40 kg.

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

Inclined Pulley

Given a 30o angle, and 2 masses each 5-kg, find the acceleration of the system, and the tension in the cord. a= 2.45m/s2. T =36.75 N

1. Derive an equation for the same inclined pulley system including friction between m1 and the ramp.

Suppose m1 = 5kg, m2 = 12-kg and the ramp angle is 20o Suppose m1 = 5kg, m2 = 12-kg and the ramp angle is 20o. Find the m that would make the system acceleration 4 m/s2. What will be the tension in the cord? The m = 0.7 T = 69 N

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