Force Systems

• Combination Systems – connected masses

• Horizontal Pulley

• Atwood’s Machine

For any force system you must sum forces.

Fnet = F = F1 + F2 …ma = F1 + F2 …

Hwk Sheet: Problems in Force 2.prb 4 – 7.

Connected Masses

What forces can you identify acting on the boxes?

What is the net force?

A constant net force, F, must accelerate the entire system’s mass.

Fnet = mtota

a = Fnet. m1 + m2 + m3 .

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.

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.

a = Fnet. m1 + m2 + m3

Find system acceleration:a = Fnet m1 + m2 + m3.

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.

1 kg

4 kg

F -T2 = m3a or F - m3a = T2.

20N - (1 kg)(2.5 m/s2) =

17.5 N

T1 = m1a = (4 kg)(2.5 m/s2) = 10 N.

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!!

• T1 = 10 N

• T2 = 17.5 N

m1a = T1.

m2a = T2 - T1.

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 2nd Law equation for each.

M1.+T.

Fn.

m1g

M2.

-T.

m2g

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 = 4.2 m/s2

T = 16.8 Nmg = 30 N, it is less than the tension.

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

Atwood’s Machine

Use wksht.

Given Atwood’s machine, m1 = 2 kg, m2 = 4 kg. Find the acceleration and tension.

Sketch the free body diagramfor 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:

m1

F2F

The net force, F1, on m1 is:

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.

100-N

1-kg

Find a for system.F2 must push the remaining 40

boxes or 40 kg.40 N.

Ignoring friction, derive an equation to solve for a and T for this system:

Begin by sketching the free body diagramWrite the equations for each boxAdd 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

Derive an equation for the same inclined pulley system including

friction.