Engineering Mechanics Formulas

1

ENGINEERING MECHANICS If the resultant is a force, R: Rx = ΣFx Ry = ΣFy ΣMo = 0 If the resultant is a couple, CR: ΣFx = 0 ΣFy = 0 ΣMo = CR

Rectangular Components of Force Q

P

 

T

Equilibrium of Rigid Bodies S Force x – component y– component P Px = P cos  Py = P sin  Q Qx = -Q sin  Qy = Q cos  T Tx = -T Ty = 0 S Sx = 0 Sy = -S Resultant of a Force System Q T

Resultant: S= R



in equilibrium if the resultant of system that acts on the body Equilibrium means that both the force and the resultant couple

O Equation of Equilibrium in Two-Dimension: ΣFx = 0 ΣFy = 0 ΣMo = 0

2

R ENGINEERING MECHANICS R y

P



A body is the force vanishes. resultant are zero.

=

 Rx

y

b

(R x )2  (R y )2

Horizontal component of resultant: Rx = ΣFx Vertical component of resultant: Ry = ΣFy Angle that the resultant makes with horizontal: Ry = R sin β Rx = R cos β

y

F a z Equation of Equilibrium in ThreeDimension: ΣFx = x0 ΣMx = 0 ΣFy = 0 ΣMy = 0 z ΣFz = 0 ΣMz = 0 To get components of force in three dimension: Fy F F F = x = = z y d x z d = distance from a to b

x

3

ENGINEERING MECHANICS (xb -x a )2  (y b -y a )2  (z b -z a ) 2

=

Reversed Effective Force

x = xb – xa y = yb – ya z = z b – za

W P

Friction

F

Friction – is the contact resistance exerted by one body upon a second body when the second body moves or tends to move past the first body. Static Friction – the two contact surfaces has no relative motion between each other. Kinetic Friction – the two contact surfaces are sliding relative to each other.

REF

Forces acting on Na body in motion: 1. Applied :P Inforce Motion 2. Weight W= mg) ( to: the right 3. Normal force : N 4. Friction : F =  N 5. Reversed Effective Force: REF = ma m = mass of body a = acceleration of the body Rectilinear Translation

F

Rectilinear Acceleration

Motion

 Impending N MotionR ( to the right )

with

Constant

motion

R = resultant of friction and normal forces = F2 + N2 tan  =   = angle of internal friction

V1

V2

Equation of motion: x V2  V1   at

V22  V12   2ax

x  V1t 

W P F

Note: The direction of friction always opposes impending sliding. The surfaces are on the verge of sliding is a condition known as impending sliding.

at2

V2 = final velocity V1 = initial velocity a = constant acceleration x = linear distance traveled

Freely N Forces acting on a body at rest : Impending Motion 1. Applied (force P to the: right ) 2. Weight : W = mg 3. Normal force : N 4. Friction : F =  N  = coefficient of friction

1 2

Falling

Bodies

(air

resistance

neglected)

Note : At the highest point the velocity is zero.

Equation of motion:

V2  V1   gt

V22  V12   2gy

y  V1t 

1 2

Equation of motion:

gt2

2 – 1 =  t 22 – 12 = 2

V2 = final velocity V1 = initial velocity g = acceleration due to gravity = 9.81 m/s2 y = vertical distance traveled

Rectilinear Diagram a

Translation

using

 = 1t 

Motion

1 2

t2

2 = final angular velocity 1 = initial angular velocity  = angular acceleration  = angular distance Curvilinear Translation

2 0

t

1 – Time Diag Acceleration

To get the velocity using acceleration – time diag:  velocity at t = 0 is the initial velocity  velocity at t = 1 is the initial velocity plus the area of a – t diag from 0 to 1

Rectangular Coordinates of Acceleration and Velocity:

v

ENGINEERING MECHANICS

at

a

v

4

V1 V2

an

V0 Velocity – Time Diag t 1 using velocity2– time diag: To 0 get the distance  

distance at t = 1 is the area of v – t diag from 0 to 1 distance at t = 2 is the area of v – t diag from 0 to 2

S2

s

Tangential acceleration : at = r Normal acceleration : an = r2 =

v2 r

Resultant acceleration : a  (at ) 2  (a n ) 2 Tangential velocity : v = r Normal velocity is zero. r = radius of the curve  = angular acceleration  = angular velocity

S1

5

ENGINEERING MECHANICS 0 Distance1– Time Diag 2 Rotation with Constant Angular Acceleration



t

Polar Coordinates of Acceleration and Velocity:

a

 

r



ar

g x2 y = x tan  2 Vo 2 cos 2 θ r – component of acceleration: a r  r  r  2

θ

–

component

of

acceleration:

a  r   2 r 

Resultant acceleration : a  (a ) 2  (a r ) 2 v

V02 sin 2  Max range of projectile, R  g 2 V sin 2  Max height of projectile, h  0 2g

where : V0 = initial velocity of projectile vr θ = the angle that V0 makes with horizontal

r

6

ENGINEERING MECHANICS

r – component of velocity: v r  r  θ – component of velocity: v  r 

Resultant velocity: v 

Work and Energy

(v ) 2  (v r ) 2

W = mg

dr where : r  dt

r   

P

d 2r dt 2 d

x

dt

V1

F V2 Work2– Energy Equation:

d 2   dt 2

Projectile Motion (air resistance neglected) y Vy2 = 0 Vx Vy1

Vo

h

 origin Vx Horizontal distance from origin at any time: x = Vx t x = Vx t = (Vocos ) t R Vertical distance from origin at any time:

1 h

N

U1-2 = T + Vg + Ve U = work done due to external forces = Force x Distance T = change in kinetic energy = T2 – T1 = ½m(V22 – V12) T2 = final kinetic energy = ½mV22 T1 = initial kinetic energy = ½mV12 Vx Vg = gravitational potential energy Vy3 =  mgh (positve if vertically upward) x Ve = elastic potential energy (due to y spring) = ½k(22 - 12) k = spring constant 2 = final deformation of spring 1 = initial deformation of spring

Impulse and Momentum V1

P

F

V2

x Impulse –NMomentum Equation: F (t) = m(V) ΣF = external forces on the body t = time interval from V1 to V2 m = mass of the body W

= g V = change in velocity = V2 – V1 Note : External forces on the body are applied and friction forces.

Prepared by: Engr. Ric O. Palma Besavilla Engg Review Center