Upload
todd-lamb
View
216
Download
0
Embed Size (px)
Citation preview
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 …
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.
1 kg
4 kg
Fnet - m3a = T2.
m1a = T1 = (4 kg)(2.5 m/s2) = 10 N.
m3a = F -T2
20N - (1 kg)(2.5 m/s2) =
17.5 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 the free body diagram each mass.
Consider any F contributing to acceleration +. Any F opposing acceleration is - .Apply 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 = 4.3 m/s2
30 N
7 kg
a = m2g m1 + m2
• (4 kg)(4.3 m/s2)
• T = 17 N
• mg = 30 N, tension less than weight.
m1a = T
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. Find the acceleration and tension.
• a = 3.3 m/s2.
• T = 26 N
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
1. Derive an equation for the same inclined pulley system including friction between m1 and the ramp.
• The = 0.7
• T = 69 N
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?
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