I taught elementary and engineering physics (algebra based) at Texas State University in San Marcos, Texas. I made this slide show really quick for a friend. I'll be adding more classical mechanics slideshows in the future.

1. ENGINEERING
Statics

2. NET FORCE
When the net force on an
object is zero, the
ACCELERATION of the object is
zero, However, the object may
still be moving IN A STRAIGHT
LINE

3. FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1
2
3
4
…. ???

4. FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1 Gravity
2 Electromagnetic
3 Strong Nuclear Force
4 Weak Nuclear Force

5. FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1 Gravity
2 Electromagnetic
3 Strong Nuclear Force
4 Weak Nuclear Force
The strong and weak nuclear forces have very short
ranges and are only effective inside the nucleus.
Therefore, you only have TWO fundamental forces to
consider! Easy, right??

6. FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1 Gravity
𝐹 = 𝐺
𝑚1𝑚2
𝑑2
In most cases, one of the two masses is the
earth (d is the radius of the earth, and Gm1/d2
is “little g” or approximately 9.8 m/s2).
Even placing an object at the top of a tall
skyscraper has little effect on gravity. The
acceleration due to gravity is usually a known
quantity given in the problem.
2 Electromagnetic

7. FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1 Gravity
𝐹 = 𝐺
𝑚1𝑚2
𝑑2
2 Electromagnetic
Coulomb force
Magnetic force
Friction
Tension
Bouyant
Normal
Spring
Intermolecular

8. FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1
2
3
4

9. FUNDAMENTAL
FORCES
How many forces are involved in a free
body diagram? Only one, possibly!
𝐹 = 𝐺
𝑚1𝑚2
𝑑2

12. WHY IS THIS SO
HARD?
1.Resolve all force
vectors into their x y
and z components
2. Σ𝐹𝑛𝑒𝑡 = 0 𝑁
3.Draw each pair of
forces involved ONE
PAIR AT A TIME
Let’s make it simpler

13. Gravity
(assume the ropes have negligible weight. Then
the force of gravity on them is zero. You may
exclude these two pairs of forces)
Intuitively, we might assume that the
force through each rope is 50N. But we
can also prove it mathematically.
𝑇1 = 𝑇2
𝑇1 + 𝑇2 = 100𝑁
A simple substitution gives us the
answer.
𝑇1 + 𝑇1 = 100𝑁
𝑇1 = 𝑇2 = 50𝑁

14. Gravity
(assume the ropes have negligible weight. Then the
force of gravity on them is zero. You may exclude
these two pairs of forces)
According to Newton’s 3rd Law of Gravity,
forces always occur in pairs.
• What object does the “reaction force”
act on?
• What is the magnitude and direction of
this “reaction force”?

15. Gravity
(assume the ropes have negligible weight. Then the force
of gravity on them is zero. You may exclude these two
pairs of forces)
Since we are only
concerned with the
forces acting on the
sign and not the
earth, we can
exclude the earth
and the reaction
force on it.

16. If the only force acting on
the sign were the weight of
the sign, there would be a
NON-ZERO net force acting on
the sign, and it would
accelerate in the direction
of the net force (in this
case, towards the ground)

17. Tension
Fortunately, there are 2
ropes supporting the weight
of the sign.
The SUM of these forces that
are acting on the sign ADD
UP TO ZERO. We know this
because the sign is NOT
ACCELERATING

18. CHECK YOUR WORK
Did we include all
relevant forces?
Suggestion – always start
with gravity
So far, we have drawn the
diagram. In the next step, we
will calculate the horizontal
and vertical components of the
tension forces in each rope.

19. 𝐹𝑛𝑒𝑡 = 0𝑁
𝐹𝑛𝑒𝑡,𝑥 = 0𝑁
𝐹𝑛𝑒𝑡,𝑥 = 0𝑁
AND

20. WHAT DO WE KNOW
𝑇1=𝑇2cosΘ
We know this because the acceleration
(and hence the net force) in the
horizontal direction is zero.
Don’t confuse these forces for
action/reaction pairs of forces!
Remember that the “reaction” forces
are acting on THE SIGN. An easy way
to remember this is to pretend the sign
were made of a stretchy material.

21. WHAT DO WE KNOW
In this problem, it is
ESSENTIAL to solve for
T2 before proceeding to
the horizontal forces.
Once you’ve done enough
of these problems,
you’ll see that each one
is a puzzle to be
solved.
𝑇1=𝑇2cosΘ