The Important Stuff The SI System
Physics is based on measurement. Measurements are made by comparisons to well–defined
standards which define the units for our measurements.
The SI system (popularly known as the metric system) is the one used in physics. Its
unit of length is the meter, its unit of time is the second and its unit of mass is the kilogram.
Other quantities in physics are derived from these. For example the unit of energy is the
joule, defined by 1 J = 1 kg·m²/s²
As a convenience in using the SI system we can associate prefixes with the basic units to
represent powers of 10.
The most commonly used prefixes are given here:
Other basic units commonly used in physics are:
Time : 1minute = 60 s 1 hour = 60 min etc.
Mass : 1 atomic mass unit = 1 u = 1.6605 × 10-27 kg
Changing Units
In all of our mathematical operations we must always write down the units and we always
treat the unit symbols as multiplicative factors. For example, if me multiply 3.0 kg by 2.0 m/s
we get
(3.0 kg) · (2.0 m/s ) = 6.0 kg·m/s
We use the same idea in changing the units in which some physical quantity is expressed.
We can multiply the original quantity by a conversion factor, i.e. a ratio of values for
which the numerator is the same thing as the denominator. The conversion factor is then
equal to 1 , and so we do not change the original quantity when we multiply by the conversion
factor.
Examples of conversion factors are:
1 min
60 s
100 cm
1m
1 year
365.25 hours
1m
3:28 ft
Density
A quantity which will be encountered in your study of liquids and solids is the density of a
sample. It is usually denoted by ρ and is defined as the ratio of mass to volume:
ρ = m/v
The SI units of density are kg/m3 but you often see it expressed in g/cm3 .
Dimensional Analysis
Every equation that we use in physics must have the same type of units on both sides of the
equals sign. Our basic unit types (dimensions) are length (L), time (T ) and mass (M).
When we do dimensional analysis we focus on the units of a physics equation without
worrying about the numerical values.
Vectors; Vector Addition
Many of the quantities we encounter in physics have both magnitude (“how much”) and
direction. These are vector quantities.
We can represent vectors graphically as arrows and then the sum of two vectors is found
(graphically) by joining the head of one to the tail of the other and then connecting head to
tail for the combination, as shown in Fig. 1.1 . The sum of two (or more) vectors is often
called the resultant.
We can add vectors in any order we want: A+B = B+A. We say that vector addition
is “commutative”.
We express vectors in component form using the unit vectors i, j and k, which each
have magnitude 1 and point along the x, y and z axes of the coordinate system, respectively.
The Important Stuff
Figure 1.1: Vector addition. (a) shows the vectors A and B to be summed. (b) shows how to perform the
sum graphically.
Figure 1.2: Addition of vectors by components (in two dimensions).
Any vector can be expressed as a sum of multiples of these basic vectors;
for example, for the vector A we would write:
A = Axi + Ayj + Azk .
Here we would say that Ax is the x component of the vector A; likewise for y and z.
In Fig. 1.2 we illustrate how we get the components for a vector which is the sum of two
other vectors. If
A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk
then
A + B = (Ax + Bx)i + (Ay + By)j + (Az + Bz)k
Once we have found the (Cartesian) component of two vectors, addition is simple; just add
the corresponding components of the two vectors to get the components of the resultant
vector.
When we multiply a vector by a scalar, the scalar multiplies each component; If A is a
vector and n is a scalar, then
cA = cAxi + cAyj + cAzk
In terms of its components, the magnitude (“length”) of a vector A (which we write as
A) is given by:
A = √ Ax2 + Ay2 + Az2
Many of our physics problems will be in two dimensions (x and y) and then we can also
represent it in polar form. If A is a two–dimensional vector and θ as the angle that A
makes with the +x axis measured counter-clockwise then we can express this vector in terms
of components Ax and Ay or in terms of its magnitude A and the angle θ. These descriptions
are related by:
Ax = Acosθ Ay = Asin θA = √ Ax2 + Ay2
tanθ =Ay/Ax
When we use tanθ =Ay/Ax to find θ from Ax and Ay we need to be careful because the inverse
tangent operation (as done on a calculator) might give an angle in the wrong quadrant; one
must think about the signs of Ax and Ay.
Multiplying Vectors
There are two ways to “multiply” two vectors together
The scalar product (or dot product) of the vectors a and b is given by :
a · b = ab cos ∅
where a is the magnitude of a, b is the magnitude of b and is the angle between a and b.
The scalar product is commutative: a · b = b · a. One can show that a · b is related to
the components of a and b by:
a · b = axbx + ayby + azbz
If two vectors are perpendicular then their scalar product is zero.
The vector product (or cross product) of vectors a and b is a vector c whose mag-
nitude is given by
c = ab sin ∅where is the smallest angle between a and b. The direction of c is perpendicular to the
plane containing a and b with its orientation given by the right–hand rule. One way
of using the right–hand rule is to let the fingers of the right hand bend (in their natural
direction!) from a to b; the direction of the thumb is the direction of c = a × b. This is
illustrated in Fig. 1.3.
The vector product is anti–commutative: a × b = −b × a.
Relations among the unit vectors for vector products are:
i × j = k j × k = i k × i = j
Worked Examples
Figure 1.3: (a) Finding the direction of A × B. Fingers of the right hand sweep from A to B in the
shortest and least painful way. The extended thumb points in the direction of C. (b) Vectors A, B and C.
The magnitude of C is C = AB sin .∅ The vector product of a and b c
by:
a × b = (aybz − azby)i + (azbx − axbz)j + (axby − aybx)k
which can be abbreviated by the notation of the determinant:
Worked Examples
Changing Units
1. The Empire State Building is 1472 ft high. Express this height in both meters
and centimeters. [FGT 1-4]
To do the first unit conversion (feet to meters), we can use the relation (see the Conversion
Factors in the back of this book):
1m = 3.281 ft
We set up the conversion factor so that “ft” cancels and leaves meters:
1472 ft = (1472 ft)(1 m/3.281 ft)
= 448.6 m .
So the height can be expressed as 448.6m. To convert this to centimeters, use:
1m = 100 cm
and get:
448.6m = (448.6m) (100 cm/1 m)
= 4.486 × 104 cm
The Empire State Building is 4.486 × 104 cm high !
2. A rectangular building lot is 100.0 ft by 150.0 ft. Determine the area of this lot
in m2. [Ser4 1-19]
The area of a rectangle is just the product of its length and width so the area of the lot
is
A = (100.0 ft)(150.0 ft) = 1.500 × 104 ft2To convert this to units of m2 we can use the relation
1 m = 3.281 ft
but the conversion factor needs to be applied twice so as to cancel “ ft2” and get “m2”. We
write:
1.500 × 104 ft2 = (1.500 × 104 ft2) · (1 m/3.281 ft2)
= 1.393 × 103 m2The area of the lot is 1.393 × 103 m2.
3. The Earth is approximately a sphere of radius 6.37 × 106 m.
(a) What is its circumference in kilometers?
(b) What is its surface area in square kilometers?
(c) What is its volume in cubic kilometers? [HRW5 1-6]
(a) The circumference of the sphere of radius R, i.e. the distance around any “great circle”
is C = 2πR. Using the given value of R we find:
C = 2πR = 2π(6.37 × 106 m) = 4.00 × 107 m .
To convert this to kilometers, use the relation 1 km = 103 m in a conversion factor:
C = 4.00 × 107 m = (4.00 × 107 m) ·(1 km/103 m )
= 4.00 × 104 km
The circumference of the Earth is 4.00 × 104 km.
(b) The surface area of a sphere of radius R is A = 4πR2.
So we get A = 4πR2 = 4π(6.37 × 106 m)2 = 5.10 × 1014 m2
Again, use 1 km = 103 m but to cancel out the units “m2 ” and replace them with “ km2 ” it
must be applied twice:
A = 5.10 × 1014 m2 = (5.10 × 1014 m2) ·(1 km/103 m2)
= 5.10 × 108 km2
The surface area of the Earth is 5.10 × 108 km2
(c)The volume of a sphere of radius R is V = 4/3 πR3.
So we get :
V = 4/3πR3 = 4/3π (6.37 × 10 6 m)3 = 1.08 × 1021 m3
Again, use 1 km = 103 m but to cancel out the units “m3” and replace them with “ km3” it
must be applied three times:
V = 1.08 × 1021 m3 = (1.08 × 1021 m3) ·(1 km/103 m )3= 1.08 × 1012 km3
The volume of the Earth is 1.08 × 10 12 km3.
4. Calculate the number of kilometers in 20.0 mi using only the following conver-sion factors:
1mi = 5280 ft, 1 ft = 12 in, 1 in = 2.54 cm, 1m = 100 cm, 1 km = 1000m.
[HRW5 1-7]
Set up the “factors of 1” as follows:
20.0 mi = (20.0mi) ·(5280 ft/1 mi).(12 in.1 ft2).(2.54 cm/1 in)·(1 m/100 cm)·(1 km/1000 m)
= 32.2 km
Setting up the “factors of 1” in this way, all of the unit symbols cancel except for km
(kilometers) which we keep as the units of the answer.
5. One gallon of paint (volume = 3.78 × 10-3 m3) covers an area of 25.0 m3. What
is the thickness of the paint on the wall? [Ser4 1-31]
We will assume that the volume which the paint occupies while it’s covering the wall is
the same as it has when it is in the can. (There are reasons why this may not be true, but
let’s just do this and proceed.)
The paint on the wall covers an area A and has a thickness ; the volume occupied is the
area time the thickness:
V = A τ .
We have V and A ; we just need to solve for τ :
τ= V/A = 3.78 × 10-3 m3/25.0 m2 = 1.51 × 10-4 m .
The thickness is 1.51 × 10-4 m. This quantity can also be expressed as 0.151 mm.
6. A certain brand of house paint claims a coverage of 460 ft2/gal .
(a) Express thisquantity in square meters per liter.
(b) Express this quantity in SI base units.
(c).What is the inverse of the original quantity, and what is its physical significance?
[HRW5 1-15]
(a) Use the following relations in forming the conversion factors: 1m = 3.28 ft and 1000 liter =
264 gal. To get proper cancellation of the units we set it up as:
460 ft2 gal = (460 ft2/gal) ·(1 m /328 ft)2· (264 gal/1000L)
= 11.3 m2/L
(b) Even though the units of the answer to part (a) are based on the metric system, they
are not made from the base units of the SI system, which are m, s, and kg. To make the
complete conversion to SI units we need to use the relation 1m3 = 1000L. Then we get:
11.3 m2/L = (11.3 m2L ) ·(1000 L / 1 m3)
= 1.13 × 104 m-1
So the coverage can also be expressed (not so meaningfully, perhaps) as 1.13 × 104 m-1.
(c) The inverse (reciprocal) of the quantity as it was originally expressed is
(460 ft2/gal)-1 = 2.17 × 10-3 gal/ft2 .
Of course when we take the reciprocal the units in the numerator and denominator also
switch places!
Now, the first expression of the quantity tells us that 460 ft2 are associated with every
gallon, that is, each gallon will provide 460 ft2 of coverage. The new expression tells us that
2.17×10-3 gal are associated with every ft2, that is, to cover one square foot of surface with
paint, one needs 2.17 × 10-3 gallons of it.7. Express the speed of light, 3.0 × 108 m/s in
(a) feet per nanosecond and
(b)millimeters per picosecond. [HRW5 1-19]
(a) For this conversion we can use the following facts:
1m = 3.28 ft and 1 ns = 10-9 s. to get:
3.0 × 108 m/s = (3.0 × 108 ms ) · (3.28 ft/1m)·(10-9 s/1 ns) = 0.98 ft/ns
In these new units, the speed of light is 0.98 ft/ns .
(b) For this conversion we can use:
1 mm = 10-3 m and 1 ps = 10-12 s
and set up the factors as follows:
3.0 × 108 m/s = (3.0 × 108 m/s ) ·(1mm/10-3 m)·( 10-12 s/1 ps )
= 3.0 × 10-1 mm/ps
In these new units, the speed of light is 3.0 × 10-1 mm/ps .
8. One molecule of water (H2O) contains two atoms of hydrogen and one atom
of oxygen. A hydrogen atom has a mass of 1.0 u and an atom of oxygen has a
mass of 16 u, approximately.
(a) What is the mass in kilograms of one molecule
of water?
(b) How many molecules of water are in the world’s oceans, which
have an estimated total mass of 1.4 × 1021 kg? [HRW5 1-33]
(a) We are given the masses of the atoms of H and O in atomic mass units; using these
values, one molecule of H2O has a mass of
m H2O = 2(1.0 u) + 16 u = 18 u
Use the relation between u (atomic mass units) and kilograms to convert this to kg:
m H2O = (18 u)(1.6605 × 10-27 kg/1 u)
= 3.0 × 10-26 kg
One water molecule has a mass of 3.0 × 10-26 kg.
(b) To get the number of molecules in all the oceans, divide the mass of all the oceans’
water by the mass of one molecule:
N = 1.4 × 1021 kg/3.0 × 10-26 kg
= 4.7 × 1046 .
. . . a large number of molecules!
Density
9. Calculate the density of a solid cube that measures 5.00 cm on each side and
has a mass of 350 g. [Ser4 1-1]
The volume of this cube is€
V = (5.00 cm) · (5.00 cm) · (5.00 cm) = 125 cm3
So from Eq. 1.1 the density of the cube is
ρ = m/V
= 350 g/125 cm3
= 2.80 g/cm3
Figure 1.4: Cross–section of copper shell in Example 11.
10. The mass of the planet Saturn is 5.64 × 1026 kg and its radius is 6.00 × 107 m. Calculate its density. [Ser4 1-2]
The planet Saturn is roughly a sphere.
(But only roughly! Actually its shape is ratherdistorted.)
Using the formula for the volume of a sphere, we find the volume of Saturn:
V = 4/3R3 = 4/3π(6.00 × 107 m)3 = 9.05 × 1023 m3
Now using the definition of density we find:
ρ =m/V
=5.64 × 1026 kg/9.05 × 1023 m3
= 6.23 × 102 kg/m3
While this answer is correct, it is useful to express the result in units of g/cm3 . Using our
conversion factors in the usual way, we get:
6.23 × 102 kg/m3 = (6.23 × 102 kg/m3 ) ·(103 g/1 kg)·(1m/100 cm)
= 0.623 g/cm3
The average density of Saturn is 0.623 g/cm3 . Interestingly, this is less than the density of water.
11. How many grams of copper are required to make a hollow spherical shell
with an inner radius of 5.70 cm and an outer radius of 5.75 cm?
The density of copper is 8.93 g/ cm3. [Ser4 1-3]
A cross–section of the copper sphere is shown in Fig. 1.4. The outer and inner radii are
noted as r2 and r1, respectively. We must find the volume of space occupied by the copper
metal ; this volume is the difference in the volumes of the two spherical surfaces:
Vcopper = V2 − V1 = 4/3πr23 − 4/3πr13 = 4/3π (r23 − r13 )
With the given values of the radii, we find:
Vcopper = 4/3π((5.75 cm)3 − (5.70 cm)3)
= 20.6 cm3
Now use the definition of density to find the mass of the copper contained in the shell:
ρ=mcopper/Vcopper
= mcopper = ρVcopper = 8.93 g/cm3(20.6 cm3) = 184 g
184 grams of copper are required to make the spherical shell of the given dimensions.
12. One cubic meter (1.00m3) of aluminum has a mass of 2.70×103 kg, and 1.00m3
of iron has a mass of 7.86 × 103 kg. Find the radius of a solid aluminum sphere
that will balance a solid iron sphere of radius 2.00 cm on an equal–arm balance.
[Ser4 1-39]
In the statement of the problem, we are given the densities of aluminum and iron:
ρAl = 2.70 × 103 kg/ m3 and ρFe = 7.86 × 103 kg/m3 .
A solid iron sphere of radius R = 2.00 cm = 2.00 × 10 -2 m has a volume
VFe = 4/3πR
= 4/3π (2.00 × 10-2 m)3
= 3.35 × 10 -5 m3so that from MFe = ρFeVFe we find the mass of the iron sphere:
MFe = ρFeVFe =( 7.86 × 103 kg/m3)(3.35 × 10−5 m3) = 2.63 × 10-1 kg
If this sphere balances one made from aluminum in an “equal–arm balance”, then they
have the same mass. So MAl = 2.63 × 10-1 kg is the mass of the aluminum sphere. FromMAl = ρAlVAl we can find its volume:
VAl =MAl/ρAl
=2.63 × 10-1 kg/2.70 × 103 kg/m3
= 9.76 × 10-5 m3
Having the volume of the sphere, we can find its radius:
VAl = 4/3R3 = R = (3/4VAl)1/3
This gives:
R =( 3(9.76 × 10-5 m3)/4π) 1/3
= 2.86 × 10-2 m = 2.86 cm
The aluminum sphere must have a radius of 2.86 cm to balance the iron sphere.
Dimensional Analysis
The period T of a simple pendulum is measured in time units and is
where ` is the length of the pendulum and g is the free–fall acceleration in units
of length divided by the square of time. Show that this equation is dimensionally
correct. [Ser4 1-14]
The period (T ) of a pendulum is the amount of time it takes to makes one full swing
back and forth. It is measured in units of time so its dimensions are represented by T .
On the right side of the equation we have the length `, whose dimensions are represented
by L. We are told that g is a length divided by the square of a time so its dimensions must
be L/T2. There is a factor of 2π on the right side, but this is a pure number and has no
units. So the dimensions of the right side are:
so that the right hand side must also have units of time. Both sides of the equation agree in
their units, which must be true for it to be a valid equation!14. The volume of an object as a function of time is calculated by V = At3+B/t,
where t is time measured in seconds and V is in cubic meters. Determine the
dimension of the constants A and B. [Ser4 1-15]
Both sides of the equation for volume must have the same dimensions, and those must
be the dimensions of volume where are L3 (SI units of m3). Since we can only add terms
with the same dimensions, each of the terms on right side of the equation (At3 and B/t)
must have the same dimensions, namely L3.
Suppose we denote the units of A by [A]. Then our comment about the dimensions of
the first term gives us:
[A]T 3 = L3
=) [A] =L3T 3
so A has dimensions L3/T 3. In the SI system, it would have units of m3/ s3.
Suppose we denote the units of B by [B]. Then our comment about the dimensions of
the second term gives us:
[B]/T = L3
=) [B] = L3T
so B has dimensions L3T . In the SI system, it would have units of m3 s.
15. Newton’s law of universal gravitation is
Here F is the force of gravity, M and m are masses, and r is a length. Force has
the SI units of kg · m/s2. What are the SI units of the constant G? [Ser4 1-17]
If we denote the dimensions of F by [F] (and the same for the other quantities) then
then dimensions of the quantities in Newton’s Law are:
What we don’t know (yet) is [G], the dimensions of G. Putting the known dimensions into
Newton’s Law, we must have:
since the dimensions must be the same on both sides. Doing some algebra with the dimen-sions, this gives:
so the dimensions of G are L3/(MT2). In the SI system, G has units of
m3/kg · s316. In quantum mechanics, the fundamental constant called Planck’s constant,
h , has dimensions of [ML2T -1]. Construct a quantity with the dimensions of length from h, a mass m, and c, the speed of light. [FGT 1-54]
The problem suggests that there is some product of powers of h, m and c which has
dimensions of length. If these powers are r, s and t, respectively, then we are looking for
values of r, s and t such that
hrmscthas dimensions of length.
What are the dimensions of this product, as written? We were given the dimensions of h, namely [ML2T -1]; the dimensions of m are M, and the dimensions of c are L
T = LT -1 (it is a speed). So the dimensions of hrmsct are:
[ML2T -1]r[M]s[LT -1]t
= Mr+s L2r+t T -r-t
where we have used the laws of combining exponents which we all remember from algebra
Now, since this is supposed to have dimensions of length, the power of L must be 1 but
the other powers are zero. This gives the equations:
r + s = 0
2r + t = 1
−r − t = 0
which is a set of three equations for three unknowns. Easy to solve!
The last of them gives r = −t. Substituting this into the second equation gives
2r + t = 2(−t) + t = −t = 1 =) t = −1
Then r = +1 and the first equation gives us s = −1. With these values, we can confidentlysay that
hrmsct = h1m-1c-1
= h/mc
has units of length.
Vectors; Vector Addition
17. (a) What is the sum in unit–vector notation of the two vectors a = 4.0i+ 3.0j
and b = −13.0i + 7.0j? (b) What are the magnitude and direction of a + b?
[HRW53-20]
(a) Summing the corresponding components of vectors a and b we find:
a + b = (4.0 − 13.0)i + (3.0 + 7.0)j
= −9.0i + 10.0j
This is the sum of the two vectors is unit–vector form.
(b) Using our results from (a), the magnitude of a + b is
|a + b| = √(−9.0)2 + (10.0)2 = 13.4
and if c = a + b points in a direction as measured from the positive x axis, then the
tangent of θ is found from
tan θ = ( cy/cx )
= −1.11
If we naively take the arctangent using a calculator, we are told:
θ = tan-1 (−1.11) = −48.0 o
which is not correct because (as shown in Fig. 1.5), with cx negative, and cy positive,
the correct angle must be in the second quadrant. The calculator was fooled because angles
which differ by multiples of 180o have the same tangent. The direction we really want is
θ = −48.00+ 180.00= 132.00
Figure 1.5: Vector c, found in Example 17. With cx = −9.0 and cy = +10.0, the direction of c is in the
second quadrant.
