Transcript Document 7764042

REVIEW FOR MID TERM 1
SI
CGS
BE
Length
Meter (m)
Centimeter
(cm)
Foot (ft)
Mass
Kilogram
(kg)
Gram (g)
Slug (sl)
Time
Second (s) Second (s)
Second
(s)
English  S.I.
• 1 inch = 2.540 cm (centimeter = 1/100
meter)
• 1mile = 1.609 km (kilometer= 1000
meters)
• 1 pound = 454 g (gram)
• 1.06 quart = 1 L (litre)
Components Of Vectors
A = Ax + Ay
Using Components To Add Vectors
Cx = Ax + Bx
Cy = Ay + By
Vav,x = (x2-x1)/(t2-t1) = Δx/Δt
SI unit: m/s
Average Acceleration
aav,x = (v2x – v1x) / (t2 - t1) = Δvx / Δt
SI unit: m/s2
Instantaneous Acceleration
ax = lim (Δvx/Δt)
Δt 0
SI unit: m/s2
Cars Accelerating or Decelerating
Constant-Acceleration Equations of Motion
Variables Related
Equation
Velocity, time,
acceleration
vx = v0x + axt
Position, time,
acceleration
x = x0 + v0xt + (½ )axt2
Velocity, position,
acceleration
vx2 = v0x2 + 2ax(x – x0) = v0x2 + 2axDx
Constant-Acceleration Equations of Motion in TwoDimensions
vx = v0x + axt
vy = v0y + ayt
x = x0 + v0xt + (½ )axt2
y = y0 + v0yt + (½ )ayt2
vx2 = v0x2 + 2ax(x – x0)
vy2 = v0y2 + 2ay(y – y0)
Equations of Motion for projectile
ax = 0m/s2
ay = -9.81m/s2
vx = v0x + axt
vy = v0y + ayt
x = x0 + v0xt + (½ )axt2
y = y0 + v0yt + (½ )ayt2
vy2 = v0y2 + 2ay(y – y0)
Determination of key items
for projectiles
• x = (vocos o)t
•  = tan-1(vy/vx)
• y = (vosin o)t - ½gt2
• vx = vocoso
• vy = vosino- gt
R = F1 + F2 + F3 + ……..= Σ F,
(resultant, and vector sum, of forces)
Rx = Σ Fx ,
Ry = Σ Fy
(components of vector sum of forces)
Once we have the components Rx and Ry, we can find the
magnitude and direction of the vector R.
Newton’s First Law – Figure 4.7
•“Objects at rest tend to
stay at rest and objects
in motion tend to stay in
motion in a straight line
unless it is forced to
change that state by
forces acting on it”
Newton’s Second Law of Motion (Vector Form)
The vector sum (resultant) of all the forces acting on an object
equals the object’s mass times its acceleration :
ΣF =
ma
The acceleration a has the same direction as the resultant force ΣF.
Newton’s Second Law of Motion (Vector Form)
The vector sum (resultant) of all the forces acting on an object
equals the object’s mass times its acceleration :
ΣF =
ma
The acceleration a has the same direction as the resultant force ΣF.
Forces are the origin of motion
Forces
Acceleration
Velocity
v= v0 + at
Position
x = x0 + v0t + ½ at2
a = F/m
y
x positive
y positive
x
y
x negative
y positive
x
y
x
x negative
y negative
y
x
x positive
y negative
y
x positive
y positive
x
??????
Drawing a FBD of forces on an object (on, not by)
1. Choose the object to analyze. Draw it as a dot.
2. What forces physically touch this object?
This object, not some other
3. What “action at a distance” forces act on the object?
Gravity is the only one for this PHYS2053
4. Draw these forces as arrows with tails at the dot (object).
5. Forces only! No accelerations, velocities, …
Get components of Newton’s 2nd Law
Choose a convenient xy coordinate system
Find the x and y components of each force in the FBD
Add the x and y components separately
In a rescue, the 70.0 police officer is suspended by
two cables, as shown in the figure below
Find the tension in the cables.
y
T2 sin480
T1
T2
T1 sin350
48o
35o
T1
cos350
T2
w
cos480
x