equilibrium & newton’s laws of motion tensional forces

12
Equilibrium & Newton’s Laws of Motion Tensional Forces

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Page 1: Equilibrium & Newton’s Laws of Motion Tensional Forces

Equilibrium & Newton’s Laws of Motion

Tensional Forces

Page 2: Equilibrium & Newton’s Laws of Motion Tensional Forces

Equilibrium

Newton’s 1st Law of Motion When the forces on an object are balance, it is

said to be in equilibrium. When an object is in equilibrium, it is not

accelerating. An object that is not accelerating is stationary or

moving at constant speed in a straight line.

You balanced forces during the force table lab.

Page 3: Equilibrium & Newton’s Laws of Motion Tensional Forces

Equilibrium in 2-Dimensions

When objects are in equilibrium: ax and ay = 0

It then follows that: ΣFx and ΣFy = 0

Because Fnet = ma and a = 0.

Page 4: Equilibrium & Newton’s Laws of Motion Tensional Forces

Ex. 1: Equilibrium

A 100 N sign is hung by two wires as seen below. What is the tension in the wires?

Physics is Fun

Fg = 100 N

FA FB

= 15°

Page 5: Equilibrium & Newton’s Laws of Motion Tensional Forces

Diagram the Problem

Physics is Fun

Fg = 100 N

FA FB

y

x

System

= 15°

y

x

FBx

FByFAy

FAx

Fg = 100 N

FA FB

Page 6: Equilibrium & Newton’s Laws of Motion Tensional Forces

State the Known & Unknown

What is known? Fg = 100N θ = 15°

What is not known? FA

FB

Page 7: Equilibrium & Newton’s Laws of Motion Tensional Forces

Perform Calculations

Isolate the x and y components separately. Since the sign is not moving, Fnet = ma = 0 in both the x and y

directions. x – direction:

-FAx + FBx = 0

-FA cosθ + FB cosθ = 0 FA cosθ = FB cosθ

= 15°

y

x

FBx

FByFAy

FAx

Fg = 100 N

FA FB

Page 8: Equilibrium & Newton’s Laws of Motion Tensional Forces

Perform Calculations

y – direction: FAy + FBy – Fg = 0 FA sinθ + FB sinθ – Fg = 0 2FA sinθ = Fg (due to symmetry, FAy = FBy) FA = (100N)/((sin15°)(2)) FA = 193 N

= 15°

y

x

FBx

FByFAy

FAx

Fg = 100 N

FA FB

Page 9: Equilibrium & Newton’s Laws of Motion Tensional Forces

Ex. 2: Equilibrium

An engine has a weight of 3150 N. The engine is positioned above the engine compartment using rope, a pulley and a ring as shown. Find the tension in each of the sections of rope

Page 10: Equilibrium & Newton’s Laws of Motion Tensional Forces

State the Known & Unknown

What is known? W = 3150 N θ1 = 10° θ2 = 80°

What is not known? T1

T2

Page 11: Equilibrium & Newton’s Laws of Motion Tensional Forces

Perform Calculations

Isolate the x and y components separately. Since the engine is not moving, Fnet = ma = 0 in both the x

and y directions. x – direction:

-T1sin θ1 + T2sin θ2 = 0

-T1sin 10° + T2sin 80° = 0 (1)

y – direction: T1cos θ1 - T2cos θ2 – W = 0

T1cos 10° - T2cos 80° - W = 0 (2)

Solve (1) for T1 and substitute into (2).

Page 12: Equilibrium & Newton’s Laws of Motion Tensional Forces

Perform Calculations (cont.)1 2

sin80.0

sin10.0T T

2 2

2

2

2

1

sin80.0cos10.0 cos80.0 0

sin10.0

:

sin80.0cos10.0 cos80.0

sin10.0

3150582

sin80.0cos10.0 cos80.0

sin10.0

sin80.0(582 ) 3300

sin10.0

T T W

Solve forT

WT

NT N

T N N