1 / 29

# Tension in a rope has two properties: * it is evenly distributed throughout the connected objects - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Tension in a rope has two properties: * it is evenly distributed throughout the connected objects' - ria-pickett

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Tensionin 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.

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)

force, NOT the direction of motion, The accelereation is constantly 9.8 m/sFy = 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 force, NOT the direction of motion, The accelereation is constantly 9.8 m/s

• Circle the 1 body you are going to analyze

• Reduce it to a point and draw a FREE BODY DIAGRAM (show each force as an arrow and label it)

• Do F=ma in one direction at a time (don’t mix x and y in the same equation)

• Repeat as often as necessary to get an equation with the one variable for which you are looking. Try other directions and other bodies.

• Solve using algebra. Harder problems may involve solving simultaneous equations

Rope Tension: Hanging Basket force, NOT the direction of motion, The accelereation is constantly 9.8 m/s

A Block & Tackle force, NOT the direction of motion, The accelereation is constantly 9.8 m/s

Equilibrium force, NOT the direction of motion, The accelereation is constantly 9.8 m/sF =0Forms Closed Triangle

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

T1

T2

T3

If the whole system accelerates at 1 m/s equilibrium.2, find the force on each object.

d c b a

1000 kg

F

800 kg

1kg

1 kg

No friction

What would happen if d and b were both eggs that broke under 2N of force?

M equilibrium.

Why must M accelerate no matter how small m is?

ICE no friction

m

Assume a perfect pulley: massless and no friction.

Why must m fall at less than 9/8 m/s2?

Derive a formula for the tension in the rope that depends only on m, M and g.

On what factors will the acceleration of the system depend?

On what factors will the tension in the rope depend?

-T equilibrium.

+N

Fy = may

+T

W - T = may

-W

+W

mg - T = may

M

ICE no friction

m

These Ts must be opposite in sign!

Fx = Max

a = (mg- T)/m

T = Max

T = M(mg- T)/m

FOIL and solve for T

T equilibrium.= M(mg- T)/m

T = (Mmg –MT)/m

T = 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)

M equilibrium.

Wood  = 0.1

m

-T

-N

Fy = may

+T

W - T = may

+W

+W

mg - T = may

Now let’s add friction. Derive a formula for the acceleration that depends only on , m, M and g.

Assume a perfect pulley

-f

Fx = Max

T - f = Max

T - uN = Max

T - uMg = Max

M equilibrium.

Wood  = 0.1

m

T - uMg = Max

mg - T = may

mg - may= T

mg -ma - uMg = Ma

mg - uMg = Ma + ma

factoring

g(m – uM) = (M + m) a

Check dimensional consistency

Check motion if M = 0, a = g

factoring

a = g(m – uM) / (M + m)

M equilibrium.

Now let’s add friction. Derive a formula for the acceleration that depends only on , m, M and g.

Wood  = 0.1

m

Assume a perfect pulley

M equilibrium.

Wood  = 0.1

m

+T

+N

Fy = may

-T

T - W = may

-W

-W

T - mg = may

Using the opposite sign convention

Assume a perfect pulley

+f

Fx = Max

f – T = Max

uN – T = Max

uMg - T = Max

M equilibrium.

Wood  = 0.1

m

uMg – T = Max

T – mg = may

T = ma + mg

uMg– (ma + mg) = Ma

uMg - ma - mg = Ma

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 equilibrium.

• 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

7 kg

3N

box.Fy = may

Equilibrium

F = ma

Fx = max

If the 3 forces were connected head to tail, they would form ………..

Equilibrium box.F =0Forms Closed Triangle

Real World box. Vector Diagram

Vector Diagram box. Resultant

Equilibrium F =0

Forms Closed Triangle