Transcript 3.3 Forces Adv B 2 MODIFIED

Tension in a rope has two properties:
* it is evenly distributed throughout the connected objects
* its always directed away from any object in the system
(you can’t push a rope)
50 N
50 N
50 N
11 lbs
Frictionless, low mass
pulleys
change
only the direction of a
tension, not its
magnitude.
Tension still is evenly
distributed throughout
the rope as it goes over
the pulley.
Reaction forces
The acceleration’s direction is NOT always the same
as the direction of it’s velocity. The acceleration is in
the direction of the net force on a body.
F = ma = W= mg
Even in projectile motion, the
acceleration follows the force,
NOT the direction of motion,
The accelereation is constantly
9.8 m/s2 downward as well,
because the only force acting is
the weight (downward)
Fy = may
Equilibrium
F = ma
Fx = max
If the 3 forces were connected head to
tail, they would form ………..
The Fool-Proof Newtonian Force
Method for Solving Problems
a) Circle the 1 body you are going to analyze
b) Reduce it to a point and draw a
FREE BODY
DIAGRAM (show each force as an arrow and label it)
c) Do
F=ma
in one direction at a time
(don’t mix x and y in the same equation)
d) Repeat as often as necessary to get an equation with the one
variable for which you are looking. Try other directions and
other bodies.
e) Solve using algebra.
Harder problems may involve
solving simultaneous equations
Rope Tension: Hanging Basket
A Block & Tackle
Equilibrium F =0
Forms Closed Triangle
Find how hard the elephant must pull
to keep the ring in equilibrium.
At this angle equilibrium is
impossible to achieve!!!!
See p. 91. The Equilibrant
must be 500N at 36.9
degrees west of South
90
135
If each block is 1 kg, find the tension in each rope.
T1
T2
T3
d
1kg
c
b
a
800 kg 1 kg 1000 kg
F
If the whole system
accelerates at 1 m/s2,
find the force on each
object.
No friction
What would happen if d and b were both eggs that broke under 2N of force?
Why must M accelerate
no matter how small m is?
M
ICE no friction
m
Assume a perfect pulley: massless and no friction.
Derive a formula for the tension
in the rope that depends only on
m, M and g.
Why must m fall at less
than 9/8 m/s2?
On what factors will the
acceleration of the system
depend?
On what factors will the
tension in the rope depend?
M
ICE no friction
m
+N
-W
-T
+T
These Ts must be
opposite in sign!
Fx = Max
T = Max
Fy = may
W - T = may
+W
mg - T = may
a = (mg- T)/m
T = M(mg- T)/m
FOIL and solve for T
T = M(mg- T)/m
FOIL
T = (Mmg –MT)/m
T = Mg –MT/m
distribute m into each term above
T + MT/m = Mg
Factor out T
T(1 +M/m) = Mg
Divide both sides by ( )
T = Mg/(1+M/m)
Clean up by multiplying by m/m
T = g Mm
(M+m)
Now let’s add friction. Derive a
formula for the acceleration that
depends only on , m, M and
g.
M
Wood  = 0.1
m
Assume a perfect pulley
-N
-T
+T
-f
Fy = may
W - T = may
+W
Fx = Max
T - f = Max
T - uN = Max
T - uMg = Max
+W
mg - T = may
T - uMg = Max
mg - T = may
mg - may= T
mg -ma - uMg = Ma
M
Wood  = 0.1
mg - uMg = Ma + ma
m
factoring
g(m – uM) = (M + m) a
Check dimensional consistency
Check motion if M = 0, a = g
factoring
a = g(m – uM) / (M + m)
Now let’s add friction. Derive a
formula for the acceleration that
depends only on , m, M and
g.
M
Wood  = 0.1
m
Assume a perfect pulley
Using the opposite
sign convention
M
Wood  = 0.1
m
Assume a perfect pulley
+N
+T
-T
+f
Fy = may
T - W = may
-W
Fx = Max
f – T = Max
uN – T = Max
uMg - T = Max
-W
T - mg = may
uMg – T = Max
T – mg = may
T = ma + mg
uMg – (ma + mg) = Ma
M
Wood  = 0.1
uMg - ma - mg = Ma
m
uMg - mg = Ma + ma
factoring
g( uM - m) = (M + m) a
Check dimensional consistency
Check motion if m = 0, a = g
factoring
a = g(uM - m) / (M + m)
Analysis of results
• a = g(m – uM) / (M + m)
• This will give + acceleration (down and to
the right) if m > uM, and a = 0 if m = uM.
• The function is undefined if m < uM since
negative a makes no physical sense; it will
never fall “up” and to the left. The reality
is a will stay at zero if a≤ 0.
Two dimensional
forces must be
summed up in the x
direction alone, the y
direction alone, then
finally connected
head to tail and added
with the pythagorean
theorem
8N
2N
3N
7 kg
Find the direction and magnitude of the
acceleration of the box.
Fy = may
Equilibrium
F = ma
Fx = max
If the 3 forces were connected head to
tail, they would form ………..
Equilibrium F =0
Forms Closed Triangle
Real World  Vector Diagram
Vector Diagram Resultant
Equilibrium F =0
Forms Closed Triangle
Inclined Plane