Transcript Force Systems - Red Hook Central School Dst

Force Systems
• Combination Systems – connected
masses
• Horizontal Pulley
• Atwood’s Machine
For any force system you
must sum forces.
Fnet = SF = 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.
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.
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 force on m1 is:
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 = 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 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.