Forces & Newton’s Laws of Motion

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Forces & Newton’s Laws of Motion. Chapter 4 (angles). Sign Problems. Equilibrium-the state where all forces are balanced and acceleration is zero. Includes both stationary and constant velocity cases. Example #15. - PowerPoint PPT Presentation

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Forces & Newton’s Laws of Motion

Chapter 4(angles)

Sign Problems

Equilibrium-the state where all forces are balanced and acceleration is zero.

Includes both stationary and constant velocity cases.

Example #15 If the sign has a mass of

15kg, what is the tension in the string? What is the force the beam exerts on the sign?

Assume equilibrium. One component at a time.

Set up equilibrium case

0

0

vertical

horizontal

F

F

cos

sin

Tx

Ty

FF

FF

Fy

Fx

FT

23

Fg

Fbeam

Equilibrium case for vertical

0sin mgFT

0 gy FF

Nmg

FT 2.376sin

0verticalF

Equilibrium case for horizontal

0horizontalF

0 Beamx FF

0cos BeamT FF

BeamT FF cos

N3.34623cos2.376

Example #16 Find the tension in

each rope.

T1T2

Break up Components

111 sinFFy

222 sinFFy

111 cosFFx

222 cosFFx

F1 F2

Fx1 Fx2

Fy1 Fy2

Fg

Horizontal Pieces

Since the sign isn’t accelerating horizontally, we can use: 021 xx FF

0coscos 2211 FF

2211 coscos FF

0 xF

Fx1 Fx2

Fy1 Fy2

Fg

Vertical Pieces

Since the sign isn’t accelerating vertically, we can use: 021 gyy FFF

0sinsin 2211 mgFF

0 F

Fx1 Fx2

Fy1 Fy2

Fg

Putting the two together

0sinsincos

cos221

1

22

mgF

F

0sinsin 2211 mgFF 2211 coscos FF

1

221 cos

cos

F

F

Putting the two together cont.

mgFF

221

1

22 sinsincos

cos

mgF

21

1

22 sinsin

cos

cos

211

2

2

sinsincoscos

mgF

Putting the two together cont.

NF 383

62sin34sin34cos62cos

8.9472

NF

F 6.214cos

cos

1

221

Example #17

What is the tension in the cables that support the sign?

Physics is

Phun!!

m = 1500 kg

40°

Example #18 – Angles with accelerationA sled of mass 30 kg is pulled with a force of 50 N at a 25 degree angle. IF the coefficient of friction between the runners and the snow is 0.1, what is the acceleration of the sled?

Example #19

A 50 kg boy is pushed on a 20 kg scooter with a force shown below at 20 degrees below the horizontal. What is the coefficient of friction that would keep the scooter from accelerating? FA=200 N

Example #20

A 50 kg girl is pushed on a 20 kg scooter with a force shown below at 20-degrees below the horizontal. What is the coefficient of friction that would keep the scooter accelerating at 1.5 m/sec2?

FA=200 N

Example #21 – Ramp Problem

A crate slides down a 37 degree incline. What is the acceleration of the crate?

Example #22

A crate slides down a 35 degree incline. What is the acceleration of the crate if the coefficient of friction between the crate and the incline is 0.2?

Example #23

A skier is accelerating down a mountain that has a 30-degree incline at 4.0 m/sec2.

A) What is the coefficient of friction?

B) How long does it take him to get to the bottom of the 0.5 km long mountain?

C) How fast is he going by the time he reaches the bottom of the mountain assuming that he goes straight down without turns?

Example #24

A crate is sliding down an incline at constant speed. If the coefficient of friction is 0.45, at what angle is the incline?

What must be for the second incline?