Transcript Force Systems - Red Hook Central School Dst

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.
m1 + m2 + m3 .
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 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.
4 kg
T1 = m1a = (4 kg)(2.5 m/s2) = 10 N.
F -T2 = m3a or
F - m3a = T2.
20N - (1 kg)(2.5 m/s2) =
1 kg
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!!
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 free body diagrams for each mass
separately. Write Newton’s 2nd Law equation for
each.
-T.
Fn.
M 1.
m 1g
m1a = T
+T.
M 2.
m 2g
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 N
mg = 30 N, it is less than the tension.
Ex: Given horizontal pulley system where m1 = 2 kg, m2 = 5 kg,
and m, the coefficient of friction between m1 and the counter is
0.35, sketch the free body diagrams for m1 and m2, 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, m1 =
2 kg, m2 = 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 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 net force, F1, on m1 is:
F2
F
m1
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 = m2a
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.
a = 1m/s2.
F2 must push the remaining 40 boxes
or 40 kg.
F = 40 kg (1m/s2.)
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
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.