Chapter 4, Part 2 Applications of Newton’s Laws · Copyright © 2010 Pearson Education, Inc....

Copyright © 2010 Pearson Education, Inc.

Chapter 4, Part 2 Applications of Newton’s

Laws


Vertical Leap

0

FN FN

FW FW FM

vi


Coupled Motions When forces are exerted on connected objects, their accelerations are the same.

If there are two objects connected by a string, and we know the force and the masses, we can find the acceleration and the tension:


Coupled Motions

If m2 = 50.0 kg and m1 = 100.0 kg, what is the acceleration? What is the tension?

T = 100.0a 50.0g – T = 50.0a

Fw1

Fw2


50.0g – 100.0a = 50.0a 50.0g = 150.0a

50.0*9.81150.0

= a = 3.27 m/s2

T = 100.0a = 327 N

(g = 9.81 m/s2)


Friction Friction is caused by surfaces that are not completely smooth:


Static Friction

This is the friction experienced by resting surfaces in contact with one another

The static frictional force depends on the normal force:

The constant µs is called the coefficient of static friction. It is specific to both surfaces.

Ff = µsFN

The static frictional force is independent of the area of contact between the surfaces.


Kinetic Friction

This is the friction experienced by surfaces sliding against one another at constant velocity

The kinetic frictional force depends on the normal force:

The constant µk is called the coefficient of kinetic friction. It is specific to both surfaces.

Ff = µkFN

The kinetic frictional force is independent of the area of contact between the surfaces.


The static frictional force keeps an object from starting to move when a force is applied. The static frictional force has a maximum value, but may take on any value from zero to the maximum,

depending on what is needed to keep the sum of forces zero.


Inclined Planes

θ

θ

FN = –FWN

FW

FWP = Fw sin θ

FWN = Fw cos θ

fs


Pulleys


Forces and the Laws of Motion

v Problem v If you use a horizontal force of 30.0N

to slide a 12.0kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?



v Solution v fk = 30.0N v m = 12.0kg

Nkk Ff µ=

N

kk F

f=µ

26.0)/8.9)(0.12(

0.302 ====smkg

Nmgf

Ff k

N

kkµ



v Problem v A force of 40.0N accelerates a 5.0kg

block at 6.0 m/s2 along a horizontal surface. How large is the frictional force?



v Solution v Fapplied = 40.0N v m = 5.0kg v a = 6.0m/s2 netappliedk FFf −=

kappliednet fFF −=

NsmkgNfk 0.10)/0.6)(0.5(0.40 2 =−=



v Problem v You are driving a 2500.0kg car at a

constant speed of 14.0m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car’s wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.0m. What is the coefficient of kinetic friction between your tires and the wet road?



v Solution v d = 25.0m v m = 2500.0kg v vi = 14.0m/s v vf = 0m/s

fk = µkFN

vf2 = vi

2 + 2as 02 = 14.02 + 2a(25.0)

a = –3.92 m/s2

fk = ma fk = 2500.0(–3.92) = –9800N

µk = 0.409800 = µk (2500*9.81)

(Use the absolute value of fk, because µk is always positive)


Homework

p. 127 Multiple Choice 13-27 (odd) pp. 132-137 49, 55, 69, 73e, 89, 95, 116

Chapter 4, Part 2 Applications of Newton’s Laws · Copyright © 2010 Pearson Education, Inc....

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Transcript of Chapter 4, Part 2 Applications of Newton’s Laws · Copyright © 2010 Pearson Education, Inc....