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ENGINEERING
Statics
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
FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1
2
3
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
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??
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
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
FUNDAMENTAL
FORCES
How many forces are involved
in a free body diagram?
1
2
3
4
FUNDAMENTAL
FORCES
How many forces are involved in a free
body diagram? Only one, possibly!
𝐹 = 𝐺
𝑚1𝑚2
𝑑2
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
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.
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
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”?
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)
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
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.
𝐹𝑛𝑒𝑡 = 0𝑁
𝐹𝑛𝑒𝑡,𝑥 = 0𝑁
𝐹𝑛𝑒𝑡,𝑦 = 0𝑁
AND
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.
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
Drawing Free Body Diagrams for Engineering Statics

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Drawing Free Body Diagrams for 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
  • 10.
  • 11.
  • 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) 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.
  • 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) 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
  • 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) 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”?
  • 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,𝑦 = 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