Transcript free body diagram fbd

FORCES
Acceleration
Velocity
v= v0 + at
Position
x = x0 + v0t + ½ at2
a=F/m
NET FORCE
FREE
BODY
DIAGRAM
FBD
From Simple to Complex
• 1) Horizontal plane
• 2) Inclined plane
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
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
??????
y
q
x
q
Horizontal plane
• 1) Horizontal force
Fa
A horizontal force Fa of 12 N is applied
To a block with mass m=6kg, on a frictionless
table. The block was originally at rest when the force
was applied. Draw a FBD and find the acceleration
of the block and its velocity after it travels 0.4m
from the origin
FBD
N
Normal force
Fa applied force
y
m
W
weight
Normal force: mg = 6kg* 9.81 m/s2 (along y)
Weight: -mg (along y)
Applied force = 12 N along x
x
The net force F : the y component is zero because the normal
Force and the weight cancel. The x component is the applied
Force. Hence Fx = 12 N and Fy = 0 N
The net force F : the y component is zero because the normal
Force and the weight cancel. The x component is the applied
Force. Hence Fx = 12 N and Fy = 0 N
Applying the second Newton/s law
F=ma
Fx = max
ax = 12N/6kg
ax = Fx/m
ax = 2 m/s2
Velocity after it travels 0.4 m from the origin:
The block was originally at rest: vx0 = 0m/s
vx2 = vx02 + 2ax(x-x0)
vx = √2ax(x-x0)
vx = 1.26 m/s
Horizontal plane
• 2) Force at an angle
q
A force Fa of 15 N making an angle of 35o from the horizontal
is applied to a block with mass m=6kg, on a frictionless table.
The block was originally at rest when the force was applied.
Draw a FBD and find the acceleration of the block and its
velocity after it travels for 5 seconds from the origin
FBD
There is no motion in the y
Direction (the block does not jump !!!)
Fy = 0 N
Hence:
Normal force N = w+Fsin(q)
Normal force
N
y
m
Fa,
Fa,x
q
Fa applied force
y
W weight
Motion along x:
Fa,x = m ax
x
ax = 15N cos (35o)/6kg
Fx = Fcos(q)
Fy = Fsin(q)
ax = 2.05 m/s2
vx = vo + axt
vx = 0m/s + (2.05m/s2)(5s)
vx= 10.25 m/s
Horizontal plane
• 3) Force at an angle + friction
q
A force Fa of 15 N making an angle of 35o from the horizontal
is applied to a block with mass m=6kg, on a table with friction
force Ff opposing the motion of the block of 5.2 N magnitude.
The block was originally at rest when the force was applied.
Draw a FBD and find the acceleration of the block and its
position after it travels for 5 seconds from the origin
FBD
There is no motion in the y
Direction (the block does not jump !!!)
Fy = 0 N
Hence:
Normal force N = w+Fsin(q)
Normal force
N
y
m
Ff
Fa,
Fa,x
q
Fa applied force
y
W weight
Motion along x:
Fa,x – Ff = m ax
x
ax = (15N cos (35o) -5.2N)/6kg
Fx = Fcos(q)
Fy = Fsin(q)
ax = 1.18 m/s2
x = xo + voxt + (1/2) axt2
x = (1/2)(1.85m/s2)(5s)2
x = 14.77 m
y
q
x
q
α = 90o - q
α
β = 90o - α
β = 90o - (90o - q)
β=q
β ???
q
Inclined plane
1) Only gravitational force
M
d= h/sin(q)
d
h
q
M =25 kg θ= 25o h= 2 m
Question: What is the velocity of the block at the bottom of the
frictionless incline?
FBD
y
N
Wx
Wy
θ
W
Wx = W sin θ
N =-Wy
q
Wy = W cos θ
x
y-component of the net force is zero!
37o
Fa
M
Ff
h
22o
A box with mass 45 kg is at rest when a force Fa (20N)making an angle
of 370 with the inclined plane. The inclined plane makes an angle of 220
with the horizontal plane. When the box is moving on the inclined
Plane, there is a friction force Ff of 5 N opposing the motion. The box was
Originally at a height h from the ground (h=4 m).
Draw the free body diagram for the box. Determine the components x, y
for all the forces acting on the box. Find the net force and acceleration
Of the box.
The conditions for a particle to
be in equilibrium
• Necessary conditions for an object to
settle into equilibrium
 SF = 0
 SFx = 0 and SFy = 0
•Both x and y forces must be considered separately.
Homework from chapter 4
• 3, 9, 14, 20, 21, 23, 30, 34, 41, 52