Download and m - University of Manitoba Physics Department

Wiley Plus Reminder!
Assignment 1
6 problems from chapters 2 and 3
Kinematics
Due Monday October 5
Before 11 pm !
Chapter 4: Forces and Newton’s Laws
•Force, mass and Newton’s three laws of motion
•Newton’s law of gravity
•Normal, friction and tension forces
•Apparent weight, free fall
•Equilibrium
Problem 4.17:
A 325 kg boat is sailing 15.0 degrees north of east at
2.00 m/s. 30 s later, it is sailing 35.0 degrees north of
east at 4.00 m/s. During this time, three forces act on the
boat:

Auxiliary engine : F1 31.0 N to 15.0 degrees north of east

Water resistance: F2 23.0 N to 15.0 degrees south of west

Wind resistance : Fw

Find the magnitude and direction of the force Fw .
Give the direction with respect to due east.
Apply Newton’s Second Law:
  

F1  F2  Fw  ma
First, find the average acceleration :
north

vi  2 m/s
  15

vf  4 m/s
  35
east
t  30 sec

vf

vi
  
v  vf  vi
Then use Newton’s second law:
north

Fw  ? 

m a  24.1 N
F1  31 N
 ?
east
 
  15
F1  F2  8N

F2  23 N
Newton’s Third Law of Motion
When you exert a force on an object, it exerts a force back
on you (otherwise you fall over).
The ball hits the block and
exerts a force F1 on it.

F1
The block exerts an equal
and opposite force F2 on
the ball.


F2  F1

F2
Action and reaction forces are equal in magnitude and opposite in direction.
Action and reaction forces act on
DIFFERENT OBJECTS!

F1
All of the forces acting on the block
All of the forces acting on
the ball

wblock

F2

wball
As action and reaction forces act on different objects, the
forces do not cancel or add … you can not apply both in the same sum
for net forces … calculate net forces only for each individual object.
Newton’s Third Law of Motion
An astronaut of mass ma = 92 kg
exerts a force P = 36 N on a
spacecraft of mass ms = 11,000 kg.
What is the acceleration of each?
Force on the spacecraft, P = 36 N to the right.
Second Law:

Acceleration of craft:

P
as 

ms
36 N
 0.0033 m/s 2
11000 kg
Reaction force of spacecraft on the astronaut is –P (Newton’s 3rd law).
Second Law:

Acceleration of astronaut:

 P  36 N
aa 

 0.39 m/s 2
ma
92 kg
The Fundamental Forces of Nature
• Strong Nuclear Force: the strongest of all.
Responsible for binding neutrons and protons into the nuclei of atoms.
Acts over only very short distances of about 10-15 m.
• Electroweak Force -a combination of:
– electromagnetic force: binds electrons to nuclei to form atoms and
molecules.
– weak nuclear force: responsible for nuclear beta (plus)-decay
(primary mechanism for energy generation in a star)
• Gravity: the weakest force of all. Significant because all matter (we
believe) generates an attractive gravitational force.
• Perhaps, a repulsive gravitational force acting at long distances
(distant galaxies appear to be moving away faster than they should
if only normal gravity acts).
Action and Reaction Forces:
also for action at a distance
Why is g=9.80 m/s2 ?
The earth exerts a
gravitational force on the
moon.
The moon exerts an equal
and opposite force on the
earth.
This means that the force must
depend on the
product of both masses:
F  ME MM
Newton’s Law of Gravitation
(deduced from observations of the motion of the planets)
Every particle in the universe exerts an attractive force on every other
particle. The gravitational force between two masses, m1 and m2, is
proportional to the product of the masses and inversely proportional to the
square of the distance between their centers.
Gm1m2
Fg 
r2
Distance between centers
of gravity
G is the universal gravitational constant:
G  6.673  10 11 Nm2 /kg 2
m1, m2 are “gravitational masses”. In all cases seen, they are
equal to the inertial masses.
Gm1m3 Gm2m3
Fg 3 

0
2
2
r13
r23
Three objects of mass m1, m2, m3 are located along a straight line. m2 is
greater then m1. The net gravitational force acting on mass m3 is zero.
Which drawing correctly represents the locations of the objects?
(Remember gravity attracts m3 toward both m1 and m2, and F ~ 1/r2 )
Newton’s Law of Gravitation
Newton’s law of universal gravitation
can be applied to extended (finite
size) objects such as planets.
The prove requires the use of
calculus (not done here).
The general equation:
Gm1m2
Fg 
r2
works as long as the mass of the
object is distributed around the
center with spherical symmetry.
In this case r is the distance
between the centers of the spheres
(not the surfaces).
Why is g=9.80 m/s2 ?
Earth
m

w
ME  5.98  10 24 kg
R
R  6.38  10 6 m
ME
What is the gravitational force (weight, w) of a mass m on the earth’s
surface?
24
GmME
m

(
5
.
98

10
kg )
11
2
2
w
 (6.637  10 Nm / kg ) 
2
R
(6.38  10 6 m )2
w  9.80  m  mg
GME
g 2
R
At the earth’s surface
Above earth’s surface, weight decreases with distance r from the
centre of the earth as 1/r2.
Weight and Gravitational Acceleration
m
Due to gravity – acceleration = a
w  mg
is downwards
gravitational mass
Newton’s second law: w
 ma
(w accelerates the mass)
inertial mass
So, the weight
and
g a
w  mg  ma
independent of mass
All objects fall equally fast in the absence of air resistance
If gravitational and inertial masses were not equal, this would not be the
case!
Problem 4.19:
A bowling ball (mass m1 = 7.2 kg, radius r1 = 0.11 m) and a
billiard ball (mass m2 = 0.38 kg, radius r2 = 0.028 m) may be
treated as uniform spheres.
What is the magnitude of the maximum gravitational force between
them?
0.38 kg
r
Problem 4.26 (6th edition):
The weight of an object is the same on planets A and B.
The mass of planet A is 60% that of planet B.
Find the ratio of the radii of the planets.
m

w
M
r
Problem 4.28:
Three uniform spheres are located at the corners of an
equilateral triangle with sides of 1.2 m. Two of the
spheres have a mass of 2.8 kg. The third sphere is
released from rest.
Considering only the gravitational forces the spheres
exert on each other, what is the magnitude of its initial
acceleration?
m2  ?


F1
F2
1.2 m 30 30 1.2 m
a
m1  2.8 kg
m1  2.8 kg