university physics: mechanics

University Physics: Mechanics

Ch5. Newton’s Law of Motion

Lecture 6

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

6/2Erwin Sitompul University Physics: Mechanics

Announcement

Next week1st half : Lecture 72nd half : Quiz 2

Material: Lecture 5 & 6

Over next week1st half : Lecture 8, Make-up Quiz 22nd half : Discussion Quiz 2


What Causes an Acceleration? Out of common experience, we know that any change in

velocity must be due to an interaction between an object (a body) and something in its surroundings.

An interaction that can cause an acceleration of a body is called a force. Force can be loosely defined as a push or pull on the body.

Sir Isaac Newton (1642—1727)

The relation between a force and the acceleration it causes was first understood by Isaac Newton.

The study of that relationship is called Newtonian mechanics.

We shall now focus on its three primary laws of motion.


Newton’s First Law: “If no force acts on a body, then the body’s velocity cannot change, that is the body cannot accelerate.”

In other words, if the body is at rest, it stays at rest. If the body is moving, it will continue to move with the same velocity (same magnitude and same direction).

Newton’s First Law


A force can cause the acceleration of a body. As the standard body, we shall use the standard kilogram. It

is assigned, exactly and by definition, a mass of 1 kg. We put the standard body on a horizontal frictionless surface

and pull the body to the right, so that it eventually experiences an acceleration of 1 m/s2.

We can now declare, as a matter of definition, that the force we are exerting on the standard body has a magnitude of 1 newton (1 N).

Force


Forces are vector quantities. They have magnitudes and directions.

Principle of Superposition for Forces: A single force with the magnitude and direction of the net force acting on a body has the same effect as all the individual forces acting together.

Newton’s First Law: (proper statement)“If no net force acts on a body (Fnet = 0), then the body’s velocity cannot change, that is the body cannot accelerate.”

Force

→


0A B CF F F netF ma

, , ,ˆ ˆi ( ) iC C x A x B xF F F F

, , 0A y B yF F

, , , 0A x B x C xF F F

Net Force Calculation using Vector Sum


Which of the following six arrangements correctly show the vector addition of forces F1 and F2 to yield the third vector, which is meant to represent their net force Fnet?

Checkpoint

→ →

→


Mass Mass is a scalar quantity. Mass is an intrinsic characteristic of a body. The mass of a body is the characteristic that relates a force

on the body to the resulting acceleration. A physical sensation of a mass can only be obtained when

we attempt to accelerate the body.


Newton’s Second Law: “The net force on a body is equal to the product of the body’s mass and its acceleration.”

The Newton’s second law in equation form

netF ma��

It the net force Fnet on a body is zero, then the body’s acceleration a is zero If the body is at rest, it stays at rest. If it is moving, it continues to move at constant velocity.

21 N (1 kg)(1 m s ) 21 kg m s

Newton’s Second Law

→

→


The vector equation Fnet = ma is equivalent to three component equation, one written for each axis of an xyz coordinate system:

net, ,x xF ma net, ,y yF ma net, .z zF ma

The acceleration component along a given axis is caused only by the sum of the force components along that same axis, and not by force components along any other axis.

Newton’s Second Law→


Free-Body Diagram

Fon book from hand

The most important step in solving problems involving Newton’s Laws is to draw the free-body diagram.

Only the forces acting on the object of interest should be included in a free-body diagram.

Fon book from earth


The system of interest is the cart

The free-body diagram of the cart

Free-Body Diagram


Example: Puck (Ice Hockey “Ball”)

Three situations in which one or two forces act on a puck that moves over frictionless ice along an x axis, in one-dimensional motion, are presented here.

The puck’s mass is m = 0.2 kg. Forces F1 and F2 are directed along the axis and have magnitudes F1 = 4 N and F2 = 2 N. Force F3 is directed at angle θ = 30° and has magnitude F3 = 1 N. In each situation, what is the acceleration of the puck?

→ →

→

1 xF ma

1x

Fa

m

4 N

0.2 kg 220m s


Example: Puck (Ice Hockey “Ball”)

The puck’s mass is m = 0.2 kg. Forces F1 and F2 are directed along the axis and have magnitudes F1 = 4 N and F2 = 2 N. Force F3 is directed at angle θ = 30° and has magnitude F3 = 1 N.

1 2 xF F ma

4 N 2 N

0.2 kgxa

210m s

3, 2x xF F ma

3, 2 xx

F Fa

m

3 2cos F F

m

(1 N)(cos30 ) 2 N

0.2 kg

25.67 m s


The Gravitational ForceThe gravitational force Fg on a body is a force that pulls on the body, directly toward the center of Earth (that is, directly down toward the ground.

gF mg��

→

gF mg

The WeightThe weight W of a body is equal to the magnitude Fg of the gravitational force on the body

W mg

Some Particular Forces


The Normal ForceWhen a body presses against a surface, the surface (even a seemingly rigid one) deforms and pushes back on the body with a normal force FN that is perpendicular to the surface.

In mathematics, normal means perpendicular.

→

N g yF F ma

According to Newton’s second law,

N (0)F mg m

NF mg

NF

NF

• Why?

Forces on a Body, Resting on a Table



FrictionThe frictional force or simply friction is a force f that resists the motion when we slide or attempt to slide a body over a surface.

Friction is directed along the surface, opposite the direction of the intended motion.

→



TensionWhen a cord (or a rope, cable, or other such object) is attached to a body and tensed, the cord pulls on the body with a force T directed away from the body.

The force is often called a tension force. The tension in the cord is the magnitude T of the force on the body.

→

→

• A cord is considered as massless and unstretchable

• A pulley is considered as massless and frictionless



Beware of High Energy Concentration


Forms of Energy


Newton’s Third Law: “When two bodies interact, the forces on the bodies from each other are always equal in magnitude an opposite in direction.”

BC CBF F

BC CBF F

(Equal Magnitudes)

(Equal Magnitudes and Opposite Directions)

CBFBCF

: The force on the book B from the box C

: The force on the box C from the book B

Newton’s Third Law


Applying Newton’s Law: Problem 1

A block S (the sliding block) with mass M =3.3 kg is free to move along a horizontal frictionless surface. It is connected by a cord that wraps over a frictionless pulley, to a second block H (the hanging block) with mass m = 2.1 kg. The cord and pulley are considered to be “massless”. The hanging block H falls as the sliding block S accelerate to the right. Find:(a) the acceleration of block S(b) the acceleration of block H(c) the tension in the cord


The Forces Acting On The Two Blocks

Free-Body Diagram for Block S and Block H

NF

NF



1

net,x xF Ma

xT Ma

net,y yF MaNF

N g 0SF F

N gSF F

net,y yF ma

gH yT F ma

The cord does not stretch, so ax of M and ay of m have the same magnitude.

of of y mx M aa a

The tension at M and the tension at m also have the same magnitude.

gHT F ma

Ma mg ma m

a gM m

1

2

2

T mg ma • Why?



(a) the acceleration of block S

(b) the acceleration of block H

(c) the tension in the cord

ma g

M m

2.1

9.83.3 2.1

23.81m s

2ˆ3.81i m sSa

2ˆ3.81j m sHa

T Ma (3.3)(3.81) 12.573 N



A cord pulls on a box of sea biscuits up along a frictionless plane inclined at θ = 30°. The box has mass m = 5 kg, and the force from the cord has magnitude T = 25 N.What is the box’s acceleration component a along the inclined plane?

Free-Body Diagram of the Box

Free-Body Diagram, Fg in components

→



net,x xF masin xT mg ma

net,y yF ma

N cos yF mg ma

N cos (0)F mg m No motion in

• y direction

N cosF mg

sinx

T mga

m

25 (5)(9.8)sin 30

5

20.1 m s• What is the meaning

of this value?

(5)(9.8)cos30 42.435 N

What is the force exerted by the plane on the box?



Homework 5: Two Boxes and A PulleyA block of mass m1 = 3.7 kg on a frictionless plane inclined at angle θ = 30° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.3 kg. What are:(a) the magnitude of the acceleration of each block,(b) the direction of the acceleration of the hanging block, and(c) the tension in the cord?

Hint: Draw the free-body diagram of m1 and m2 first.


Homework 5

Two blocks of mass 3.5 kg and 8.0 kg are connected by a massless string that passes over a frictionless pulley (see figure below). The inclines are frictionless.Find:(a) the magnitude of the acceleration of each block; and(b) the tension in the string.Hint: Draw the free-body diagram of the two blocks.

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university physics: mechanics

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standard body

body fnet

mechanicsa force

single force

net force fnet

net force acts

bodys acceleration

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