CE 1111 Module 5
Mathematics of Engineering
Mathematics
Saint Louis University (SLU)
67 pag.
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CE 1111:
MATHEMATICS
OF ENGINEERING
ENGR. JANRY GARCIA
ECE Faculty
School of Engineering and Architecture
Saint Louis University
jvgarcia@slu.edu.ph
09176206177
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Applicatio
ns of
Linear
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“
TLO 5: Solve worded problems
involving linear equations..
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STUDENT
+ENGAGE
ACTIVITIES:
+EXPLORE
+EVALUATE
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EXPLAIN
Define terms, Solution of a Linear Equation in One
Unknown, Literal Equation, Systems of Two Linear
Equations, Systems of Three Linear Equations
Methods of Solving Systems of Two Linear Equations,
How to Work with Word Problems
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DEFINITION
 A statement of equality
between two quantities
algebraic expressions
is called an
EQUATION.
 The two quantities are
called MEMBERS of
the equation.
 In an equation, the
literal numbers whose
values are being
determined are called
the UNKNOWNS or
VARIABLES.
 A SOLUTION of an
equation is a set of
values of the unknowns
satisfying the equation.
 When an equation
involves only one
unknown, each solution
is called a ROOT of the
equation.
 EXTRANEOUS ROOTS
are values of the
unknown obtained from
derived equations which
do not satisfy the
original equation.
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DEFINITION
 SOLVING EQUATIONS
– to solve an equation
is to find all of its
solutions.
 A Literal Equation is
an equation involving
at least 2 letters or
unknowns. Solving for
a particular variable of
a literal equation is
separating the
unknown on one side of
the equation and the
rest of the letters on
the other side.
 If equal members be
added to, subtracted
from, multiplied by,
divided by the same
numbers, the results are
equal. (Note: Division of
zero is excluded)
 If a = b, then and ,
provided a, b, and m are
not zero.
 LINEAR EQUATION –
an equation of the first
degree.
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SOLUTION OF A LINEAR
EQUATION IN ONE
UNKNOWN
+ Divide both
+ Transpose all terms
involving the unknown to
one member and all
other terms to the other
member. Combine terms
in the unknown.
sides by the
coefficient of
the unknown.
+ To check,
substitute the
result in the
original
equation.
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Examples
 3x – 5 = 7 – x
 4 (x – 3) = 2x – 2
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Examples
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Examples
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Literal Equations:
Examples
 Solve for L, r and a of the
equation
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Literal Equations:
Examples
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+
SYSTEMS OF TWO LINEAR
EQUATIONS
Consider
two linear equations in two unknowns x and y
This system of two linear equations in two unknowns is called
the simultaneous linear equations in two unknowns. A
pair of numbers x and y which satisfies both equations is
called simultaneous solution of the given equations.
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THREE METHODS OF SOLVING
SYSTEMS OF TWO LINEAR
GRAPHICAL METHOD
Example
EQUATIONS
Consistent equations have a
unique solution and the graph
is a pair of intersecting lines.
The point of intersection of the
lines is the solution set of the
given pair of equations.
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
SYSTEMS OF TWO LINEAR
EQUATIONS
SUBSTITUTION
METHOD
a. Solve one equation for x in terms of y or for y in terms of x.
Substitute this expression for that variable into the other
equation
b. Solve the resulting equation in one variable
c. Substitute the solution from step b into either original
equation to find the value of the other variable.
d. Check the solution in both of the given equations.
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1
Example -----eq’n
-----eq’n 2
From eq’n 1:
-----subst. in eq’n 2
-----subst. in 1
Therefore:
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
SYSTEMS OF TWO LINEAR
EQUATIONS
ELIMINATION
METHOD
a. To eliminate one variable, the coefficients in that unknown
must be numerically equal. If the signs of the equal
coefficients are unlike, add the equations; if like, subtract
them.
b. If the coefficients in one variable are not numerically
equal, multiply the given equation or equations by a
constant/s so that the resulting equations would have the
same numerical coefficients in that variable.
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Examples
-----eq’n 1
-----eq’n 2
-----eq’n 1
-----eq’n 2
Multiply eq’n 2 by 2
-----eq’n 1
Therefore: x=3, y=2
Therefore: x=2, y= -1
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
SYSTEMS OF THREE LINEAR
A EQUATIONS
linear equation in three variables is of the form
where a, b, c, and d are constants and a, b, and c are not
equal to zero
 To solve a system of three linear equations in three
unknowns, eliminate one unknown from a chosen pair of
equations and then eliminate THE SAME unknown from
another pair of equations. Then solve simultaneously.
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Examples
-----substitute in eq’n 4
From 1 and 2 eliminate x by
multiplying eq’n 2 by 2 then
subtract:
Substitute y and z in eq’n 1:
From 2 and 3 eliminate x by
multiplying eq’n 2 by 3 then
subtract:
Therefore :
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
SYSTEMS OF THREE LINEAR
CRAMER’S
RULE – another method of solving systems of
EQUATIONS
linear equations.
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Examples
Therefore :
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1.
HOW TO WORK WITH WORD
Read the problem all the way through quickly to see what kind of word
problem
it is and what it is about.
PROBLEMS
2. Look for a question at the end of the problem. This is often a good way to
start to find what you are solving for. Sometimes two or three things need
to be found.
3. Start every problem with "let x=something" (we generally use x for the
unknown). You let x equal to what you are trying to find. What you are
trying to find is usually stated in the question at the end of the problem.
This is called the unknown. You must show and label what stands for in
your problem, or your equation has no meaning.
4. If you have to find more than one quantity or unknown, try to determine
the smallest unknown. This unknown is often the one to let x be equal to.
5. Go back and read the problem again. This time read it one piece at a time.
Simple problems generally have two statements. One statement helps you
set up the unknowns, and the other gives you the equation information.
Translate the problem from words to symbols one piece at a time.
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Statements in Worded
Problems
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WORD PROBLEMS AND
APPLICATIONS
 NUMBER-RELATED PROBLEMS
 GEOMETRY-RELATED PROBLEMS
 DIGIT PROBLEMS
 AGE PROBLEMS
 MONEY-RELATED PROBLEMS
 SIMPLE INTEREST PROBLEMS
 MIXTURE PROBLEMS
 CLOCK PROBLEMS
 WORK PROBLEMS
 MOTION PROBLEMS
 BALANCE PROBLEMS
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Examples: NUMBER-RELATED
The sum of two numbers is 19. The larger
PROBLEMS
number
is one more than twice the smaller
number. What are the numbers?
Let: x- be the smaller number
y-be the larger number
Therefore the numbers are: 6
and 13
From 1:
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Examples
Ten less than four times a certain number is
14. Determine the number.
Let: x- be the number
Therefore the 3rd integer is 15
Three times the first of three consecutive
odd integers is three more than twice the
third. Find the third integer.
Let n- be the integer
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Examples
Find two consecutive even integers such that the square of the
larger is 44 greater than the square of the smaller integer.
Let: x- be the smaller even number
x+2- be the larger even number
Therefore the numbers are: 10 and 12
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Examples
The denominator of a certain fraction is three more than twice the
numerator. If 7 is added to both terms of the fraction, the resulting
fraction is 3/5. Find the original fraction.
Let: n- be the numerator
d-be the denominator
Therefore the original fraction is 5/13
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Examples:
GEOMETRY-RELATED
The base of an isosceles triangle is 6 cm shorter than its equal sides. If
the perimeter is 87 cm, find the length of the base.
PROBLEMS
Let: b- be the base of the isosceles triangle
s-be the equal sides of the isosceles triangle
P-is the perimeter of the isosceles triangle
Therefore the length of the base is 25 cm.
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Examples
The width of a rectangle is 9 cm. The length is 1 cm shorter than the
diagonal. Find the length of the diagonal.
Let: W- be the Width of the rectangle
L-be the Length of the rectangle
d-be the diagonal of the rectangle
Therefore the length of the diagonal is 41 cm
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Examples
Three circles are tangent externally. The distances between their centers are
58 m, 63 m, and 81 m. Find the radius of the largest circle.
***If the 3 circles are externally tangential to each other, then it means that
the line connecting their centers passes through the tangential point. Which
means that the distance between their centers is anyway equal to the sum of
their radii.
Let: x- be the radius of the smallest circle
y- be the radius of the bigger circle
z- be the radius of the largest circle
Subtract eq’ns 1 and 2:
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Examples
Add eq’ns 3 and 4:
Therefore, the radius of the largest circle is 43 m.
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Examples
The hypotenuse of a right triangle is 25
If c=17, the values of the base and
cm longer than one leg and 32 cm longer the height are negative
respectively, so it is to be rejected
than the other leg. Find the area of the
and consider c=97.
triangle.
Let c- be the hypotenuse of the right
triangle
b-the base of the right triangle
h-be the height of the right triangle
A- be the area of the triangle
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Examples
A vertical pole was broken by the wind. The
By Pythagorean Theorem:
upper part, still attached, reached a point on
the level ground 15 feet from the base. If the
upper part is 9 feet longer than the lower part,
how tall was the pole?
Let x- be the length of the lower part
x+9- be the upper part
h-be the height of the pole=x+x+9=2x+9
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Examples:
DIGIT
PROBLEMS
In a two-digit number, the units digit is 3 greater than the tens digit.
Find the number if it is 4 times as large as the sum of its digits.
Let u- be the units digit
t- be the tens digit
Therefore, the 2-digit number is 36
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Examples
The sum of the digits of a three-digit
number is 14. The hundreds digit
being 4 times the units digit. If 594 is
subtracted from the number, the order
of the digits will be reversed. Find the
number.
Let u- be the units digit
t- be the tens digit
h-be the hundreds digit
Therefore, the 3-digit number is 842
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Examples
A number is less than 100 and its tens digit is 2 more than its
units digit. If the number with the digits reversed is subtracted
from the original number, the remainder is 3 times the sum of
the digits. Find the number.
Let u- be the units digit
t- be the tens digit
Therefore, the number is 42
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Examples:
AGE PROBLEMS
Two years ago, a father was four times as old as his son. In 3 years, the
father will only be three times as old as his son. How old was the father
when his son was born?
Let x- be the son’s age now
2 years
ago
Now
In 3 years
Son
x-2
x
x+3
Father
4(x-2)
4(x-2)+2
3(x+3)
Father’s age now is 42, therefore Father’s age when the sun was born is 30
years old.
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Examples
Two years ago, the ratio of the ages of
a boy and a girl was 3 is to 4. In 8
years, their ages will be in the ratio 4 is
to 5. Find the sum of their ages now.
Let x- be the boy’s age now
Y-be the girls/ age now
2 years
ago
Now
In 8 years
Boy
3x
x
4(x+8)
Girl
4y
y
5(y+8)
Therefore, the sum of their ages
is 56.
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Examples
Richard is 5 years older than Paul. The product of their ages is 21
years less than 15 times the sum of their ages. How old is Paul
now?
Let x- be Richard’s age now
y-be Paul’s age now
Therefore, Paul’s age is 27 years old.
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Examples
Noel is 5 years older than Dennis and 10 years younger than Hilda.
In 8 years, their combined ages will be 65. How old is Noel?
Let x- be Noel’s age now
y-be Dennis’ age now
z-be Hilda’s now
Therefore, Noel’s age is 12 years old.
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Examples
The ages of the mother and her daughter are 45 and 5 respectively.
How many years will the mother be three times as old as her
daughter?
Let x- be the number of years this event will happen
Therefore, in 15 years, the mother’s age will be 3 times as old as
her daughter
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Examples:
MONEY-RELATED
Dennis has P50, P100, and
P200 paper
bills amounting to P20,300. Delio
PROBLEMS
borrowed 14 pieces of P50 bills so that
the total amounts of P50 and P100 bills
are equal. Donalyn also borrowed 11
pieces of P200 bills so that the total
amounts of the remaining P200 and
P100 are equal. How many paper bills
were left?
Let x- be the number of P50 bills
y- be the number of P100 bills
z- be the number of P200 bills
Remaining Bills:
Therefore, 203 bills remaining
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Examples
A couple does not wish to spend more than P700 for dinner at a
restaurant. If a sales tax of 6% is added to the bill and they plan to
tip 15% after the tax has been added, what is the most they can
spend for the meal?
Let x- be the most money they can spend for the meal
Therefore, P574.24 is the most they can spend for the meal.
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Examples
A man is paid P200 for each day he works and forfeits P50 for each
day he is idle. At the end of 25 days he nets P4500. How many days
did he work?
Let x- be the number of days the man worked
25-x-the number of days he was idle
Therefore, the man worked for 13 days
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Examples
Kryse has 56 coins consisting of 10-peso and 5-peso coins. If she has
a total of Php 440.00, how many coins of each type does she have?
Let x- be the number of 5 peso coins
y- be the number of 5 peso coins
Eliminate y by multiplying eq’n 1 by 10 then subtract eq’n 2.
Therefore, Kryse has 24 pieces 5-peso coin and 32 pieces of 10-peso
coin
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Examples: SIMPLE INTEREST
PROBLEMS
A retired government employee invested
P25,000 of his retirement pay at 16% per
annum. He found another investment
opportunity at 20% per annum where he
invested the rest of his retirement pay. If he
realized a total yearly income of 19% on his
two investments, what has his retirement
pay?
Let x- the total retirement pay
y- future amount for the combined retirement
pay
P1-principal for 16%; P2-principal for 20%
F1- future for 16%; F2-future for 20 %
I1-interest for 16%; I2-interest for 20%
r1=16%; r2=20% and r=final interest
rate=19%
t-is the time for which the principal amount is
given to someone
Therefore his total retirement pay is
P100, 000
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Examples
A man, speculating in real estate, invested
P300,000 in two land developments. One
investment yields 12% and the yearly income
from this is P3,600 more than the other
investment which yields 18%. How much was From 1:
the total annual income?
Let I1-interest for 12%; I2-interest for 18%
I-total yield=I1+I2
r1 and r2-rate of interest
P1 and P2-principal for 12% and 18%
respectively
t-is the time for which the principal amount is
given to someone
I1=I2+3600
I=Prt
Therefore, the man’s total yield is P42,
480
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Examples
A sum of money of simple interest amounted
to P7,800 after five years and P8,840 after
Equate eq’n 1 and 2:
nine years. Find the amount of the money
after 15 years.
Let I-interest amount
F-future amount
P-present amount
t-is the time for which the principal amount is
given to someone
From 1:
Therefore, the amount of money after 15
years is P10, 400.
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Examples: MIXTURE PROBLEMS
A 50 mL 40% acid solution is
added to a 50 mL 30% acid
solution. What will be the
concentration of the resulting
mixture?
How much of a 90% solution of
insect spray must a farmer add
to 200 cc of a 40% insect spray
to make a 50% solution of insect
spray?
Let x-be the concentration of the
resulting mixture
Let x-be the amount or volume
of the insect spray solution
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Examples
A 700-pound alloy containing 50%
tin and 25% lead is to be added with
amounts of pure tin and pure lead to
make an alloy which is 60% tin and
20% lead. Determine how much
pure tin must be added.
Let x-pounds of tin to be added
y-pounds of lead to be added
Eliminate x in eq’ns 1 and 2 by multiplying
eq’n 2 by and a subtract from eq’n 1
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Examples
How much water must be evaporated
from 12 liters of 3 kg/L salt solution until
the concentration becomes 3.60 kg/L?
Let x- be the amount of water to be evaporated
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Examples: CLOCK PROBLEMS
At what time between 4:00 PM and
5:00 PM will the hands of the clock
be coincident?
So at 4:00 the minute hand has
covered 0 degrees and hour hand has
covered 120 degrees. Now let time after which
these two coincide be x min.
Let x- be the time the hands of the clock
coincide
So hour hand covers 120+0.5x up to that time
and minute hand covers 6x degrees up to that
time when they coincide the angles should be
same, so
We know that the minute hand of a clock
covers 360 in 60 min or 6 in 1 minute and the
hour hand of a clock
covers 360 in 12 hrs. or 30 in 1 hour
or .5 degree in 1 min.
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Examples
In how many minutes after 2 o’clock will the hands of the clock
extend in opposite directions for the first time?
Let x- be the movement of the minute hand or long hand for an
analog clock
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Examples
In how many minutes after 10
o’clock will the hands of the clock
be perpendicular for the first time?
Let x- be the time the hands of the clock
perpendicular for the first time
We know that the minute hand of a clock
covers 360 in 60 min or 6 in 1 minute and
the hour hand of a clock covers 360 in 12
hrs. or 30 in 1 hour or 0.5 in 1 minute. So
the total angle movement per minute is 60.5=5.5 per minute.
At exactly 10 am, angle between hour hand and
minute hand is 60°. So further 30° movement is
needed for both hands to be perpendicular to
each other.
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Examples
Find the angle between the hands
of the clock at 3:43 PM.
Let x- be the angle between the hands of
the clock
We know that the minute hand of a clock
covers 360 in 60 min or 6 in 1 minute and
the hour hand of a clock covers 360 in 12
hrs. or 30 in 1 hour or 0.5 in 1 minute.
***Angle from hour hand to minute hand
at 3:43
At 3:43, the hour hand has moved 223 out of
720 possible times from the top of the clock.
223 times 0.5 is 111.5.
At 3:43, the minute hand has moved 43 out of
60 possible times from the top of the clock. 43
times 6 is 258.
***Angle from minute hand to hour
hand at 3:43
The angle from the minute hand to the
hour hand is simply 360 degrees minus
the degrees from the hour hand to minute
hand that we calculated.
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Examples: WORK PROBLEMS
Popoy can dig a ditch in 4 hours.
John can dig the same in 3 hours.
How long would it take them to
dig it together?
A can do a piece of work alone in 30
days, B in 20 days, and C in 60
days. If they work together, how
many days would it take them to
finish the work?
Let x- be the amount of work done together
Let x- be the amount of work done
together
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Examples
If Paolo can do his chores in ¾ of
an hour, and if Miriam and Paolo
together can do them in ½ of an
hour, how long will it take Miriam
to do it alone?
Let x- be the amount of work Miriam to do
the work alone
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Examples: MOTION
PROBLEMS
A plane flew at 20/27 of its usual
If the speed of a racing car is
increased by 20 kph, it will cover
in 7 hours the same distance it
can ordinarily cover in 8 hours.
What is its ordinary speed?
Let x- be the ordinary speed
rate in a 3,000 km course due to
inclement weather; thereby taking
an additional 1-1/2 hours to its
usual time required for the trip.
What is the usual trip of the
plane?
Let x- be the rate
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Examples
Car A can travel around a circular
track in 120 seconds while car B in
80 seconds. If they started from
the same point but travel in
opposite direction at the same
time, in how many seconds will
they meet for the first time?
Method 2:
Let x- be the time they will meet for the
first time
Method 1: Let L be the distance of the
track
To get the distance, get LCM: (120,80)
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Examples: BALANCE PROBLEMS
From eq’n 1:
A and B together weigh 242
pounds. They balance when A is
seated 5 feet from the fulcrum on
one side of a lever and B is seated
6 feet from the fulcrum on the
other side. Find the weight of
each.
Let x- be the weight of A
y-be the weight of B
From eq’n 1:
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Examples:
A uniform beam 15 feet long
weighs 3 pounds per linear foot.
At what point should it be
supported by a fulcrum if a weight
of 25 pounds on one end is
balanced by a weight of 65
pounds on the other end?
The moment of the force (which is
weight) at the (15-x)-th part of the
beam is
The condition of equilibrium is equality
of these two moments of force
Let x- be the distance from the end point
loaded by the weight od 25lbs to the
fulcrum
Then the distance from the other end
to the fulcrum is (15-x) feet.
The moment of the force (which is
weight) at the (x)-th part of the beam
is
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Examples:
John, Peter and Jane weigh 80, 60
and 50 lbs respectively. John sits 3
ft., Peter sits 5 ft. and Jane sits 6
ft. from the fulcrum on the same
side. How far must their 200 lb.
father sits from the fulcrum in
order to balance them?
Let x = distance of father from fulcrum
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ONLINE ACTIVITIES:
Do NOT forget to
answer all
requirements posted
in the Google
Classroom and submit
it on time.
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Thanks!
Any questions?
You can find me at:
+ jvgarcia@slu.edu.ph
+ 09176206177
+ Messenger
+ Private message in the Google
class
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