phy2053, lecture 8: applying newton’s laws · 2013-09-23 · phy2053, lecture 8: applying...

PHY2053, Lecture 8: Applying Newton’s Laws



Applying Newton’s Laws● Book-Proposed Strategy:● Decide which object will have Newton’s 2nd law applied● Identify all the external forces acting on that object● Draw a FBD to show all the forces acting on the object● Choose a coordinate system.

If the direction of the net force is known, choose axes so that the net force is along one of the axes

● Find the net force by adding the forces as vectors● Use Newton’s second law to relate the net force to the

acceleration● Relate the acceleration to the change in the velocity

vector during a time interval of interest

2


Friction● Force due to imperfections of surfaces in contact● For a pair of surfaces, the frictional force depends (only)

on the normal force between the surfaces:

Frictionalforce

Normal force

Coefficient of friction

● To a good approximation, doesn’t depend on size or shape of surface, only the materials in contact

● Always opposite to the direction of motion● Static friction – objects not moving w.r.t. each other● Kinetic friction – one object is being dragged across

the (surface of the) other


Problem: Pushing a crate at an angle

● A crate with a mass of 100 kg is being pushed forward across a horizontal surface.

● The coefficient of friction between the crate and the surface is 0.2.

● The force is being applied to the crate at a 30º angle with respect to the vertical axis.

● What is the minimum force at which the crate will move fromits prone position? [use g = 10 m/s2]

4

30∘

F

100 kg


Solution:

5


Solution, continued:

6


m2m1

Tension Force● “Ideal String” concept:● massless, infinitely thin, fixed length● withstands any force applied to it without breaking● Tension (T) is the magnitude of the force propagating

through the string

“String with tension T”

PHY2053, Lecture 8: Applying Newton’s Laws 8

Problem: Edge Pulley

100 kg

50 kg

Consider the system on the picture. The 100 kg block is initially held in place. The coefficient of friction between the 100 kg block and the surface is 0.2. Ignore the mass of the pulley and string. [g = 10 m/s2].

(a) If the 100 kg block is released, does the system move?

(b) If so, what is the acceleration of the system?


Solution 4,

if m2g < fk ,the system does

not move


H-ITT 1: PulleyConsider the depicted system which is released from a fixed position. The two weights have masses 120 kg and 80 kg, respectively. Assume g = 10 m/s2. The 120 kg weight will start moving:

a) downward at 4 m/s2

b) downward at 2 m/s2

c) the weight is not movingd) upward at 4 m/s2

e) upward at 2 m/s2

120 kg

80 kg


Solution to HITT #1


Concept: Inertial Reference FramesNewton’s first law defines a set of reference frames for which the observed physics will be the same. All systems which move with constant velocities with respect to each other will record the same velocity changes when performing measurements:

An inertial system is an idealized construction. In our calculations, we will assume that a fixed observer on the Earth’s surface is in an inertial reference frame.


Concept: Apparent Weight Classic application of non-inertial reference frames is an elevator accelerating or slowing down. Objects inside the elevator appear to be gaining or losing weight depending on the direction of the acceleration.

x

y

Na

mg

Newton’s 2nd law for the object:

Apparent Weight:


Problem: Apparent Weight

In an elevator that has a constant acceleration upward, Luke is standing on a scale. The scale reads 960 N.When Luke picks up a 20 kg box, the scale reads 1200 N. The acceleration of the elevator stays the same.

a) Find the acceleration of the elevatorb) Find Luke’s weight


Solution:

15


H-ITT 2: Apparent Weight


H-ITT 2: Apparent WeightLuke’s mass is 100 kg. He is standing on a scale in an elevator that is slowing down at 5 m/s2 as it reaches the top floor of a building. Assume g = 10 m/s2. The scale will show an apparent weight of:

✔

1)0 N2)500 N3)750 N4)1000 N5)1500 N


Solution to HITT #2

18


H-ITT 2: Apparent WeightLuke has a mass of 100 kg. He is standing on a scale in an elevator which is accelerating as it heads for the top floor of a building at 5 m/s2. Assume g = 10 m/s2. The scale will show an apparent weight of:

✔

1)0 N2)500 N3)750 N4)1000 N5)1500 N


Solution to HITT #2

20


Concept: Air Resistance● air resistance: the force with which air opposes

movement of objects through air● air resistance is velocity dependent - the higher the

velocity of the object, the higher the air resistance● a simple model is that in which Fair resist = αvobject

● consequence: terminal velocity during freefall

mg

Fair

v < vterm

mg

Fair

v > vterm

Fair

mg

v = vterm

phy2053, lecture 8: applying newton’s laws · 2013-09-23 · phy2053, lecture 8: applying...

Documents