tension in a rope has two properties: * it is evenly distributed throughout the connected objects
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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. - PowerPoint PPT PresentationTRANSCRIPT
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
11 lbs
50 N50 N
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)
Equilibrium
F = ma Fx = max
Fy = may
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 =0Forms Closed Triangle
Find how hard the elephant must pull to keep the ring in equilibrium.
90
135
At this angle equilibrium is impossible to achieve!!!!
See p. 91. The Equilibrant must be 500N at 36.9 degrees west of South
T1
T2
T3
If each block is 1 kg, find the tension in each rope.
800 kg1 kg
1000 kg1kg
F
No friction
If the whole system accelerates at 1 m/s2, find the force on each object.
What would happen if d and b were both eggs that broke under 2N of force?
d c b a
ICE no friction
M
m
Assume a perfect pulley: massless and no friction.
Why must M accelerate no matter how small m is?
Why must m fall at less than 9/8 m/s2?
On what factors will the acceleration of the system depend?
Derive a formula for the tension in the rope that depends only on m, M and g.
On what factors will the tension in the rope depend?
ICE no friction
M
m
+T+N
-W
Fx = Max
-T
+W
Fy = may
W - T = may
mg - T = may
T = Max
These Ts must be opposite in sign!
a = (mg- T)/m
T = M(mg- T)/m FOIL and solve for T
T = M(mg- T)/m T = (Mmg –MT)/mT = Mg –MT/m
T + MT/m = Mg
T(1 +M/m) = Mg
T = Mg/(1+M/m)
T = g Mm
FOIL
distribute m into each term above
Factor out T
Divide both sides by ( )
Clean up by multiplying by m/m
(M+m)
Wood = 0.1
M
mAssume a perfect pulley
Now let’s add friction. Derive a formula for the acceleration that depends only on , m, M and g.
-f
+T-N
+W
Fx = Max
T - f = Max
T - uN = Max
T - uMg = Max
-T
+W
Fy = may
W - T = may
mg - T = may
T - uMg = Max mg - T = may
mg - may= T
mg -ma - uMg = Ma
mg - uMg = Ma + ma
factoring
g(m – uM) = (M + m) a
factoring
a = g(m – uM) / (M + m)
Check dimensional consistency
Check motion if M = 0, a = g
Wood = 0.1
M
m
Wood = 0.1
M
mAssume a perfect pulley
Now let’s add friction. Derive a formula for the acceleration that depends only on , m, M and g.
Wood = 0.1
M
mAssume a perfect pulley
Using the opposite sign convention
+f
-T+N
-W
Fx = Max
f – T = Max
uN – T = Max
uMg - T = Max
+T
-W
Fy = may
T - W = may
T - mg = may
uMg – T = Max T – mg = may
T = ma + mg
uMg – (ma + mg) = Ma
uMg - ma - mg = Ma
factoring
g( uM - m) = (M + m) a
factoring
a = g(uM - m) / (M + m)
Check dimensional consistency
Check motion if m = 0, a = g
Wood = 0.1
M
m
uMg - mg = Ma + ma
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
7 kg2N
3N
8N
Find the direction and magnitude of the acceleration of the box.
Equilibrium
F = ma Fx = max
Fy = may
If the 3 forces were connected head to tail, they would form ………..
Equilibrium F =0Forms Closed Triangle
Real World Vector Diagram
Vector Diagram Resultant
Equilibrium F =0
Forms Closed Triangle
Inclined Plane