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