1. Introduction to StaticsB C D E F G H A P O N M L K J I
Statics is the branch of mechanics concerned with the analysis of loads (forces, torque/moment) on physical systems in static equilibrium which is when a system is either at rest, or its center of mass moves at constant velocity.

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3. Units of Measure physical quantity dimension SI system U.S. Customary
Most units of measure can be broken down into the basic units of length (L), mass (M) and time (t).
physical
quantity
dimension
SI system
U.S. Customary
system
length 𝐿
𝑚 𝑜𝑟 𝑚𝑚
1𝑚= 1,000𝑚𝑚
𝑖𝑛 𝑜𝑟 𝑓𝑡
12𝑖𝑛=1𝑓𝑡
area
𝐿 2
𝑚2 𝑜𝑟 𝑚𝑚2
1 𝑚 2 = 1,000 𝑚𝑚 2 =1,000,000 𝑚𝑚 2
𝑖𝑛2 𝑜𝑟 𝑓𝑡2
1 𝑓𝑡 2 = (12 𝑖𝑛) 2 =144 𝑖𝑛 2
volume
𝐿 3
𝑚3 𝑜𝑟 𝑚𝑚3
1 𝑚 3 = (1,000𝑚𝑚) 3 = 𝑚𝑚 3
𝑖𝑛3 𝑜𝑟 𝑓𝑡3
1 𝑓𝑡 3 = (12 𝑖𝑛) 3 =1728 𝑖𝑛 3
1 𝑦𝑑 3 = (3𝑓𝑡) 3 =27 𝑓𝑡 3

4. Units of Measure physical quantity dimension SI system U.S. Customary
force
𝐹=𝑚∙𝑎
𝑀∙𝐿 𝑡 2
𝑁= 𝑘𝑔∙𝑚 𝑠 2
𝑘𝑁=1,000𝑁
𝑙𝑏 𝑙𝑏𝑓
1𝑘𝑖𝑝=1,000𝑙𝑏
pressure or stress
= 𝑓𝑜𝑟𝑐𝑒 𝑎𝑟𝑒𝑎
𝑀∙𝐿 𝑡 2 ∙ 1 𝐿 2 = 𝑀 𝐿∙ 𝑡 2
𝑁 𝑚 2 =𝑃𝑎
𝑘𝑃𝑎=1,000𝑃𝑎
M𝑃𝑎=1,000,000𝑃𝑎
𝑙𝑏 𝑖𝑛 2 =𝑝𝑠𝑖
energy
=𝑓𝑜𝑟𝑐𝑒∙𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑀∙𝐿 𝑡 2 ∙𝐿 = 𝑀∙ 𝐿 2 𝑡 2
𝑁∙𝑚=𝐽 (𝐽𝑜𝑢𝑙𝑒)
𝑓𝑡∙𝑙𝑏
power
= 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑖𝑚𝑒
𝑀∙ 𝐿 2 𝑡 2 ∙ 1 𝑡 = 𝑀∙ 𝐿 2 𝑡 3
𝑁∙𝑚 𝑠 = 𝐽 𝑠 =𝑊 (𝑊𝑎𝑡𝑡)
𝑓𝑡∙𝑙𝑏 𝑠
550 𝑓𝑡∙𝑙𝑏 𝑠 =ℎ𝑝
L = length M = mass t=time

5. Three Characteristics of Forcesq 𝐹 x y
denotes a vector
Magnitude (F)
how hard you are pushing or pulling
graphically represented by the length of the arrow
Direction (q)
usually defined by an angle measured relative to a coordinate system
Point of Application

6. Breaking a Force into x and y Components
Recall trigonometric functions: q 𝐹 x y
sin 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐹 𝑦
𝐹 𝑦 α
sin 𝜃 = 𝐹 𝑦 𝐹
cos 𝜃 = 𝐹 𝑥 𝐹
𝐹 𝑥
𝐹 𝑦 =𝐹∙ sin 𝜃
𝐹 𝑥 =𝐹∙ cos 𝜃
How will the expressions change for 𝐹 𝑥 and 𝐹 𝑦 if you were given α instead of θ?

7. Resultants 𝑅 𝐹 2 (moved) 𝐹 2 𝐹 1
How can we find the net effect of two or more forces that act on a body? 𝑦 𝑅
𝐹 2 (moved)
𝐹 2
𝐹 1
Head to Tail 𝜃 𝑥
Point of Application
Geometric Solution:
Arrange applied forces head to tail
Draw a vector starting at the tail of the first vector to the head of the last vector
The new vector is the resultant

8. Resultants 𝑅 𝐹 2 𝐹 1 𝑅 𝑥 =Σ 𝐹 𝑥 𝑅 𝑦 =Σ 𝐹 𝑦 Generally…
𝑅 𝑥 =Σ 𝐹 𝑥
𝑅 𝑦 =Σ 𝐹 𝑦
Generally…
How can we find the net effect of two or more forces that act on a body? 𝑦
Algebraic Solution:
To obtain the components of the resultant force, sum the corresponding components of each force. 𝑅
𝑅 𝑦
𝐹 2𝑦
𝐹 2
𝑅 𝑥 = 𝐹 1𝑥 + 𝐹 2𝑥 = 𝐹 1 cos 𝜃 𝐹 2 cos 𝜃 2
𝐹 2𝑦
𝐹 1
𝑅 𝑦 = 𝐹 1𝑦 + 𝐹 2𝑦 = 𝐹 1 sin 𝜃 𝐹 2 sin 𝜃 2
𝐹 1𝑦 𝜃
To find the resultant, R, of the Rx and Ry components, use Pythagorean Theorem
𝜃 2
𝜃 1 𝑥
𝐹 2𝑥
𝑅 𝑥
𝐹 2𝑥
𝐹 1𝑥
𝑅= 𝑅 𝑥 𝑅 𝑦 2
To find the angle of the resultant, use the tangent function
𝜃= 𝑡𝑎𝑛 −1 𝑅 𝑦 𝑅 𝑥

9. Class Problem: Two ants are hitched to a beetle and pull with the forces shown:
Find the x and y components of the force exerted by ant 1.
Find the x and y components of the force exerted by ant 2.
Find the x component of the resultant force of the ants.
Find the y component of the resultant force of the ants.
Find the resultant force of the ants.
What angle does the resultant force make
with the x-axis? 𝑦
Ant 1 = 𝑁
Solution: a.
= 𝑁
𝐹 1𝑥 = 𝐹 1 cos 30° = 𝑁∙cos⁡(30°)
30° 𝑥
= 𝑁
𝐹 1𝑦 = 𝐹 1 sin 30° = 𝑁∙sin⁡(30°)
50°
Beetle b.
= 𝑁
𝐹 2𝑥 = 𝐹 2 cos 50° = 𝑁∙cos⁡(50°)
=− 𝑁
𝐹 2𝑦 = 𝐹 2 sin 50° = 𝑁∙sin⁡(50°)
Ant 2 = 𝑁
Negative since the force acts in the –y direction. You also know the result should be negative because sin⁡(50°) in QIV is negative.
When accounting for the sign of a force, it is most intuitive to simply look at the force diagram and set forces to + or – based on their coordinate direction.

10. Solution: d.
= 𝑁
𝐹 1 = 𝑁
𝑅 𝑥 = 𝐹 1𝑥 + 𝐹 2𝑥 = 𝑁 𝑁
𝐹 1𝑥 = 𝑁 𝑦
𝑅 𝑦 = 𝐹 1𝑦 + 𝐹 2𝑦 = 𝑁− 𝑁
=− 𝑁
𝐹 1𝑦 = 𝑁
Ant 1 = 1.5g
𝑅= 𝑅 𝑥 𝑅 𝑦 2 = ( 𝑁 ) 2 + (− 𝑁 ) 2 e.
= 𝑁
𝜃= tan −1 𝑅 𝑦 𝑅 𝑥 = tan −1 − 𝑁 𝑁 f.
=−16.83°
30°
−16.83° 𝑥
50°
Beetle
Ant 2 = 2g
𝐹 2 = 𝑁
𝐹 2𝑥 = 𝑁
𝐹 2𝑦 =− 𝑁

11. Class Problem: Three water skiers are pulled by a boat as shown
Class Problem: Three water skiers are pulled by a boat as shown. Determine
The resultant force of the water skiers.
The direction of the resultant measured counterclockwise from the positive x-axis.
Draw the resultant force and label the computed angle on an x-y coordinate system.
𝐹 1𝑥 =−140𝑙𝑏∙𝑠𝑖𝑛⁡(40°)
=−89.99𝑙𝑏
𝐹 2𝑥 =−100𝑙𝑏∙𝑐𝑜𝑠⁡(15°)
=−96.59𝑙𝑏
𝐹 1𝑦 =140𝑙𝑏∙𝑐𝑜𝑠⁡(40°)
=107.25𝑙𝑏
𝐹 2𝑦 =100𝑙𝑏∙𝑠𝑖𝑛⁡(15°)
=25.88𝑙𝑏
𝐹 3𝑥 =−200𝑙𝑏∙𝑐𝑜𝑠⁡(65°)
=−84.52𝑙𝑏
𝐹 3𝑦 =−200𝑙𝑏∙𝑠𝑖𝑛⁡(65°)
=−181.26𝑙𝑏
𝛽=190.07°
=−89.99𝑙𝑏−96.59𝑙𝑏−84.52𝑙𝑏
=−271.1𝑙𝑏
𝑅 𝑥 = 𝐹 1𝑥 + 𝐹 2𝑥 + 𝐹 3𝑥
=107.25𝑙𝑏+25.88𝑙𝑏−181.26𝑙𝑏
=−48.13𝑙𝑏
𝑅 𝑦 = 𝐹 1𝑦 + 𝐹 2𝑦 + 𝐹 3𝑦
𝑅= 𝑅 𝑥 𝑅 𝑦 2
= −271.1𝑙𝑏 −48.13𝑙𝑏 2
=275.34𝑙𝑏
𝛽= tan −1 −48.13𝑙𝑏 −271.1𝑙𝑏
=10.07°
in QIII, so ° measured CCW from positive x-axis