Physics 101: Lecture 02
Forces: Equilibrium Examples
Today’s lecture will cover Textbook
Sections 2.1-2.7
Phys 101 URL:
http://courses.physics.illinois.edu/phys101/
Read the course web page!
Physics 101: Lecture 2, Pg 1
Overview
Last Lecture
Newton’s Laws of Motion
» FIRST LAW: Inertia
» SECOND LAW: Fnet = ma
» THIRD LAW: Action/reaction pairs
M Earth m æ M Earth ö
Gravity W =G 2 = mç G 2 ÷
rEarth è rEarth ø
= mg (near Earth’s surface!)
Today
Forces as Vectors
Free Body Diagrams to Determine Fnet
» Draw coordinate axes, each direction is independent.
» Identify/draw all force vectors
Friction: kinetic f = mkN; static f ≤ ms N
Contact Forces – Springs and Tension
Physics 101: Lecture 2, Pg 2
Forces as Vectors
Last lecture we calculated the force of gravity on a book
(i.e. its WEIGHT):
Calculate the gravitational force on a 3 kg book held 1 meter
above the surface of the earth.
W = G MEarth m / rEarth2
= (6.7x10-11 m3 / (kg s2)) (6x1024 kg) (3 kg)/ (6.4x106 + 1)2 m2
= 29.4 kg m/s2 = 29.4 N
We missed something: The direction! W
is different than
W
Physics 101: Lecture 2, Pg 3
Forces as Vectors
A quantity which has both magnitude and direction is
called a VECTOR; FORCES are VECTORS
Usually drawn as an arrow pointing in the proper
direction, where the length indicates the magnitude
W1 W2 = 2W1
= W1 + W1
This is an example of VECTOR ADDITION: to add
vectors, you place them head to tail, and draw the
RESULTANT from the start of the first to the end of the
last A B C
+ =
Physics 101: Lecture 2, Pg 4
Another Example of a Force:
Tension
Tension in an Ideal String, T:
Direction is parallel to string (only pulls) T
Magnitude of tension is equal everywhere. T
m
Now we are ready to do some physics!
QUESTION:
We suspend a mass m = 5 kg from the ceiling
using a string. What is the tension in the string?
Physics 101: Lecture 2, Pg 5
Newton’s 2nd Law and Equilibrium Systems
We suspend a mass m = 5 kg from the ceiling
using a string. What is the tension in the string?
Every single one of these problems is done the same way!
Step 1: Draw a simple picture (called a Free Body Diagram), and
label your axes! +y
-y
Step 2: Identify and draw all force vectors Weight, W Tension, T
Step 3: Use your drawing to write down Newton’s 2nd law
FNet = ma In equilibrium, everything is balanced! a=0
T-W =0 T = W = mg = (5 kg)*(9.8 m/s2) = 49 N
Physics 101: Lecture 2, Pg 6
Checkpoint!
•What does scale 1 read?
• A) 225 N B) 550 N C) 1100 N
T=W
T=W
The magnitude of tension in a ideal string is equal everywhere.
Physics 101: Lecture 2, Pg 7
Tension ACT
Two boxes are connected by a string over a
frictionless pulley. In equilibrium, box 2 is lower
than box 1. Compare the weight of the two boxes.
A) Box 1 is heavier
B) Box 2 is heavier
C) They have the same weight
Step 1 – Draw! +y
1
+y
Step 2 – Forces! T T
Step 3 – Newton’s 2nd! 2
1 2
FNet = m a m1g m2g
1) T - m1 g = 0 -y -y
2) T - m2 g = 0
Physics 101: Lecture 2, Pg 8
Another Force Example: Springs
Force exerted by a spring is directly proportional
to its displacement x (stretched or compressed).
Fspring = -k x
Example: When a 5 kg mass is suspended from a spring,
the spring stretches x1 = 8 cm. If it is hung by two
identical springs, they will stretch x2 = +y
A) 4 cm B) 8 cm C) 16 cm
S1 S 2
1 Spring 2 Springs
S1 - W = 0 S1 + S2 - W = 0 W
-y
S1 = W kx2 + kx2 = 2kx2 = W = mg
x2 = mg/(2k) = (5kg)*(9.8m/s2)/
kx1 = mg
(2*612.5N)
k = mg/x1 = 612.5 N/m So: x2 = 4 cm.
Physics 101: Lecture 2, Pg 9
2 Dimensional Equilibrium!
Calculate force of hand to keep a book sliding at
constant speed (i.e. a = 0), if the mass of the book
is 1 Kg, ms=.84 and mk=.75
We do exactly the same thing as before, except in
both x and y directions! +y
Normal
Step 1 – Draw! friction
Step 2 – Forces! -x +x
W
Step 3 – Newton’s 2nd (FNet = ma)! Hand
-y
Treat x and y independently!
FNet, y = N – W = may = 0
FNet, x = H – f = max = 0
Physics
This is what we want!
Physics 101: Lecture 2, Pg 10
Calculate force of hand to keep the book sliding at a constant speed
(i.e. a = 0), if the mass of the book is 1 Kg, ms = .84 and mk=.75.
FNet, y = N – W = 0 N=W
FNet, x = H – f = 0 H=f
Magnitude of frictional force is proportional to the normal force and
always opposes motion!
fkinetic = mk N mk coefficient of Kinetic (sliding) friction
fstatic ≤ ms N ms coefficient of Static (stationary) friction
H = f = mkN = mkW = mkmg
= (0.75)*(1 kg)*(9.8 m/s2)
H = 7.35 N
Physics 101: Lecture 2, Pg 11
Forces in 2 Dimensions: Ramp
Calculate tension in the rope necessary to keep the
5 kg block from sliding down a frictionless incline
of 20 degrees.
Step 1 - Draw!
You should draw axes parallel and perpendicular to motion!
Step 2 - Forces! N
T
W
Weight is not in x or y direction!
Need to DECOMPOSE it!
q Physics 101: Lecture 2, Pg 12
Vector Decomposition
N
Wy q
T Wx W
qW
y q
W Wx
Now: Step 3 – Newton’s 2nd! Split W into COMPONENTS
parallel to axes
Note that
W Wy Wx
Using Trig: Wx = W sin q
q Wy = W cos q
Physics 101: Lecture 2, Pg 13
Calculate force necessary to keep the 5 kg block from
sliding down a frictionless incline of 20 degrees.
N Now: Step 3 – Newton’s 2nd!
T Wx x direction:
Wy Fnet, x = max
System is in equilibrium (a = 0)!
Wx = W sin q
Fnet, x = 0
Wy = W cos q
Wx - T = 0
T = Wx = W sin q
mg sin q
= (5kg)(9.8m/s2) sin(20o)
q T = 16.8 N
Physics 101: Lecture 2, Pg 14
Normal Force ACT
What is the normal force of the ramp on the block?
A) N > mg B) N = mg C) N < mg
y direction:
Wx = W sin q N Fnet, y = may
Wy = W cos q T Wx Equilibrium (a = 0)!
Wy Fnet, y = 0
N - Wy = 0
q Physics 101: Lecture 2, Pg 15
Summary
Contact Force: Tension
Force parallel to string
Always Pulls, tension equal everywhere
Contact Force: Spring
Can push or pull, force proportional to displacement
F = k x
Contact Force: Friction
Static and kinetic
Magnitude of frictional force is proportional to N
Two Dimensional Examples
Choose coordinate system; choose wisely!
Analyze each direction independentlyPhysics 101: Lecture 2, Pg 16
Force at Angle Example
A person is pushing a 15 kg block across a floor with mk= 0.4 at a
constant speed. If she is pushing down at an angle of 25 degrees, what
is the magnitude of her force on the block?
x- direction: FNet, x = max Combine:
Px – f = P cos(q) – f = 0 (P cos(q) / m) – mg – P sin(q) = 0
P cos(q) – m N = 0 P ( cos(q) / m - sin(q)) = mg
N = P cos(q) / m P = m g / ( cos(q)/m – sin(q))
y- direction: FNet, y = may P = 80 N
N – W – Py = N – W – P sin(q) = 0
Normal
N – mg – P sin(q) = 0
Pushing y
q x
q Friction
Weight
Physics 101: Lecture 2, Pg 17
Tension Example:
Determine the force exerted by the hand to
suspend the 45 kg mass as shown in the picture.
A) 220 N B) 440 N C) 660 N
D) 880 N E) 1100 N y
x
T T
FNet = m a
T+T–W=0
W
•Remember the magnitude of the tension is the
same everywhere along the rope! Physics 101: Lecture 2, Pg 18
Tension ACT II
Determine the force exerted by the ceiling to
suspend pulley holding the 45 kg mass as
shown in the picture.
y
A) 220 N B) 440 N C) 660 N
Fc x
D) 880 N E) 1100 N
SF = m a
Fc -T - T – T = 0 T
•Remember the magnitude of the tension is the
same everywhere along the rope! Physics 101: Lecture 2, Pg 19
