Engineering Mechanics

Engineering Mechanics is the science which considers the effects of forces on rigid bodies. The subject is divided into two parts.

  1. Statics – the effects and distribution of force on rigid bodies which are and remain at rest.
  2. Dynamics – consider the motion of rigid bodies caused by the forces acting upon them.

Engineering Mechanics
Force System – a force system

– is any arrangement where two or more forces act on a body or on a group of related bodies

Resultant of two concurrent coplanar forces

Engineering Mechanics

Resultant of two or more Concurrent coplanar

Engineering Mechanics

Resultant of Concurrent Forces in Space

Engineering Mechanics

Given the three concurrent forces which through (1, -3, 4) and the indicated points
F1 = 150N (5, -6, 2)
F2 = 340N (4, 0, -3)
F3 = 280N (-1, 2, 6)

Determine the magnitude of resultant force.

Solution: Resolution of Forces

Engineering Mechanics

Finding the Differential Equation from a General Solution

To find the differential equation when the general solution is given, differentiate the general solution, differentiate the derived solution etc. until the number of derived equation is equal to the number of independent arbitrary constant finally eliminate the constants from the derived equations.

Finding the Differential Equation from a General Solution Examples

Example # 1:

Find the differential equation of X^2 + Y^2 = CX

Differential Equation

Example # 2:

Find the Differential Equation of ( X – C1 )^2 + C2y = C3

Differential Equation

Example # 3:

Find the differential Equation of all lines through the origin is

y = mx

m = y/x

Differential Equation

Example # 4:

Find the differential equation of all circles through (0,0) and (2,0)


The standard equation of a circle is (x-h)^2 + (y-k)^2  =  1-2

Differential Equation

Linear First and Second Order Differential Equation

Linear First and Second Order Differential Equation

Linear First Oder Differential Equation

The equation in the form dy/dx + Py = Q where P and Q are functions of x only is called Linear Differential Equations since y and its derivatives are of the first degree.

The solution for dy/dx + Py = Q is obtained by multiplying throughout by an Integrating factor Linear First and Second Differential Equation  to become Linear First and Second Differential Equation

Example: Solve the equation dy +4xy dx = 2xdx


Rearranging:  dy/dx + 4xy = 2x

then P = 4x and Q = 2x


Linear Second Order Differential Equation

Equation in the form a (d^2y/dx^2) + b(dy/dx) + cy = 0 where a, b, and c are constants, is called a linear second order differential equation with constant coefficient

Setting D = d/dx and D^2 = d^2/dx. The following procedures may be followed.

  1. write the equation in D – operator form (aD^2 + bD + C) y=0, substitute m for D and solve the auxiliary equation am^2 + bm + c=0 for m

A. If the roots are real and different (b^2 > 4ac) say Linear First and Second Differential Equation.

Then the general solution is Linear First and Second Differential Equation

B. If the roots are real and equal Linear First and Second Differential Equation twice the general solution is Linear First and Second Differential Equation

C. If the roots are imaginary (b^2 – 4ac) Say Linear First and Second Differential Equation

the general solution is Linear First and Second Differential Equation

Example:  Solve the equation 2(d^2y/dx^2) + 5(dy/dx) – 3y = 0


Writing D – operator form: (2D^2  + 5D – 3) y = 0

Substituting m for D gives the auxiliary equation 2m^2 + 5m – 3 = 0 which can be factored as (2m – 1) (m + 3) and the roots are m = ½ and m = -3

Since the roots are real and different the general solution is  Linear First and Second Differential Equation

then the general solution is Linear First and Second Differential Equation

Linear First and Second Differential Equation

How to Solve Differential Equation with Example

How to Solve Differential Equation with Example

Differential Equation

– are equation that contain differential coefficients.


 Differential Equation Example 1

– classified according to the highest derivative that occurs in Them the differential equation dy/dx = 12x is a first order differential equation d^2y/dx^2 + 4dy/dx – 3y = 0 is a second order differential equation. A solution to a differential equation that contains one or more arbitrary constant of integration is called general solution. When additional information is Given so that these constants may be calculated the particular solution of differential equation is obtained.

Variable Separable

A differential equations can be of type dy/dx = f(x)solved by direct integration by writing it in the form dy = f(x) dx


Solve the differential equation dy/dx = 2x + Sin 3x

Solution: dy = (2x + Sin 3x) dx

y = x2 – 1/3 Cos 3x + c  general solution

Differential equation of type dy/dx = f(y) can be solved by direct integrating by writing it in the form.

dx = dy/f(x)


Solve the equation (y^2 – 1) dy/dx = 3y Given that y = 1 when x = 13/6

Differential Equation Example 2

Differential Equation of type

dy/dx = f(x)  g(y) can be solve by Direct Integration by writing it in the form dy/g(y) = f(x)dx

Differential Equation of type dq/dt = KQ the general solution of an equation of the form dq/dt = KQ is Q = Ce^kt . where C is constant


solve the equation dy/dx = 3y

here we have  Q = y   dQ = dy                     then

t = x       k = 3                          y = Ce^3x


Obtain the differential equation of the family of straight lines with slope and y intercept equal


The standard equation of the line in slope – intercept form is y = mx + b.

Since the slope and y intercept are equal m = b = c then y = cx + c

isolating constant and differentiate

Differential Equation Example 3

Teaching and Learning of Japanese and Mandarin Chinese

Acquiring a second language is hard enough; learning a foreign tongue is even harder. This difficulty is particularly more pronounced in learning East Asian languages such as Japanese and Mandarin Chinese whose grammatical structure, orthography, and phonology seem too exotic for English users.

Thankfully, there is Bikol which, in my own classroom experience, provides the following learning aids:
a) prior knowledge upon which the students can build new learning;
b) cognitive landmark that may guide the students in navigating that complicated maze called foreign language acquisition and transitioning them from the familiar to the unknown; and
c) basis for analogy or comparison that may enhance comprehension.


Teaching Japanese Phonetics
It is a well known fact that Japanese cannot produce the /l/ sound. This is not a racial predisposition. Perhaps, the Japanese are not able to develop the ability to perceive, hence, articulate the /l/ sound because such sound in nonexistent in their native language. By nature, such ability would find no usefulness.

And then there is this mistaken notion that the Japanese language is replete with the /r/ sound. Actually “r” is a mere symbol used to represent a sound that is midway between /r/ and /l/. This sound, although described as “midway between /r/ and /l/,” is actually closer to /d/.

The following texts that describe the Japanese /r/ and give suggestions on how it should be properly articulated were verbatimly lifted from authoritative sources and given to the students for them to understand.

“r—a sound peculiar to Japanese pronounced with the tip of the tongue moving midway in the mouth but not rolled. If the tongue is given slightly more tension, this sound easily becomes “d.” It is like neither “r” nor “l” in English but is sort of between the two, like the Spanish “r” in “pero.”

“/r/ is probably the most difficult sound for English-speaking students. This is neither /r/, /l/, nor /d/ in English, but may be most close to /d/. In Japanese it is necessary to distinguish /r/ and /d/.In both cases the tongue touches somewhere and comes off quickly and decisively. The difference between these /r/ and /d/ is that of the position of contact.

/r/ against the alveolar ridge with the very tip of the tongue
/d/ against the teeth with somewhere more front of the tongue”

“I put my foot in my mouth again today. When I introduced Miss Winters to my Japanese friends I meant to say Edo bungaku o kenkyuu shite imasu (She’s studying the literature of the Ed period). But my “d” sounded like an “r.” So what I actually said was “She’s studying pornographic literature” [Ero bungaku o kenkyuu shite imasu].”

“I then remembered what Sensee told me about the pronunciation of the “d” sound; I should have pressed the tip of the tongue against the upper teeth ridge.”

Teaching Japanese Orthography

Filipino students find the Japanese hiragana (cursive script) and katakana (angular script) too intimidating. They find such writing systems completely foreign. But using baybayin as springboard may somehow ease the intimidation and melt the resistance against learning them.

“Ra” is formed simply by adding a diacritic to “da.” Such process may also be used as springboard in teaching the transformation of basic kana characters into sonant and semisonant characters.


Japanese Adjectives
Japanese adjectives are of two kinds: i-adjectives and na-adjectives.
I-adjectives, considered as true adjectives, are native to the Japanese language. They are so called because they end in character い(„i‟).
Example: oishii (delicious)
atsui (hot)
atatakai (warm)
When used as noun modifiers, i-adjectives behave like English adjectives, that is, they precede the nouns that they modify. For example:
oishii tabemono (delicious food)
atsui kōhii (hot coffee)
atatakai tenki (warm weather)
Na-adjectives are those that originated from Chinese and other foreign languages.
Example: shizuka (peaceful)
nigiyaka (lively)
shinsetsu (kind)
They are called na-adjectives because they need particle “na” when modifying nouns. For instance:
shizuka na machi (peaceful town)
nigiyaka na pātii (lively party)
shinsetsu na sensei (kind teacher)
Most students could not grasp the sense of the English translation because they could not account for the missing “na.” But translating the Japanese phrases into Bikol does the trick.
shizuka na machi (matoninong na banwa)
nigiyaka na pātii (maribok na party)
shinsetsu na sensei (maboot na maestra)
By employing Japanese-Bikol analogy, the Bikol translation enhances comprehension.

Mandarin Chinese Basic Sentence Pattern
The first sentence learned by a student of Mandarin Chinese is usually “Nǐ (you) hăo (fine).” It‟s literal translation is “you are fine,” but its equivalent expression in English is “Hi.”
“Nǐ hăo” may be expanded to “Nǐ hăo ma?” whose literal translation is “you are fine?” but is equivalent in English to “How are you?”
The formulaic response is “Wŏ (I) hăo” (I am fine) or “Wŏ hĕn (very) hăo” (I am very fine).
“Wŏ hăo” is the basic—being the simplest—sentence pattern in Mandarin Chinese. Ironically, it is its simplicity that makes it hard to comprehend. English users are told that a sentence may not be complete without a verb. But “Wŏ hăo” is a complete, albeit verbless sentence.
In Mandarin Chinese, when the complement is an adjective, a linking verb is not necessary, as in:
Wŏ hăo. (I am fine.)
Hànyŭ (Mandarin Chinese) nán (difficult). Mandarin Chinese is difficult.
But the complement is a noun, the linking verb “shì” is needed, as in:
Wŏ shì (am) Fēilǜbīnrén (Filipino). (I am [a] Filipino.)
Tā (she) shì lăoshī (teacher). (She is a teacher.)



How to Solve Projectile Motion Problems

Projectile Motion is a form of motion where a particle (called a projectile) is thrown obliquely near the earth’s surface, and it moves along a curved path under the action of gravity. The path followed by a projectile motion is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory after which these is no interference apart from gravity.

These are the formulas to solve Projectile Problems.

Projectile Motion Formulas

Example of Projectile Motion Problems with Solution:

1. A stone is projected from the ground with a velocity of 15 m/s at an angle of 30 degree with the horizontal ground. How high in meter will it rise? Use g = 9.817 m/s/s.


Projectile Motion 1

Projectile Motion Solution

2. Former Bob and his horse Flash jump a 100m wide canyon. The flight through the air takes 5s. Find the initial velocity (That is, find the magnitude and the angle of the initial velocity)

1st You need to solve the vertical using the formula V = Vo  +  gt. Substitute the gravity and the time.

Projectile Motion Formula Vertical Solution 

2nd Solve the horizontal, use the formula x = Vox times the time.

Projectile Motion Horizontal Solution

 Use Pythagorean Theorem formula to solve initial Velocity and the Angle.

Answer: 31.3 m/s for the Initial Velocity , and 50.8 degree for the Angle.

3. A rock is thrown horizontally off a 100m cliff. It lands 95m away. At what speed was it thrown?

Projectile Motion 2


4. A ball is thrown from a tower 30m high above the ground with a velocity of 300 m/s directed at 20 degree from the horizontal. How long will the ball hit the ground?

Projectile Motion with Problems and Solution

5. A Projectile is shot from edge of a cliff 125m above ground level with an Initial speed of 65 m/s at an angle of 37 degree above the horizontal. Determine the magnitude and the direction of the velocity at the maximum height.

Answer: The Projectile’s velocity at its max height is completely horizontal, and this is the same as the initial horizontal velocity.

Vf = Vix

     = 65Cos 37

     = 51.9 m/s (forward/horizontal)