Coplanar Vectors and Cartesian Representa1on Op#mist – The glass is half full. Pessimist – The glass if half empty. Engineer – The glass is overdesigned. Review ¢  Two forces can be combined using the parallelogram law to form a resultant ¢  A resultant can be broken up into its components using the geometry of the system and some trig 2
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Forces and vector components ¢  Two forces act on the hook. Determine the magnitude of the resultant force. To solve the problem we utilized the parallelogram law. We moved the 500 N
force, maintaining its orientation, until the tail of the 500 N force was
positioned on the head of the 200 N force. We then solved for the angle
between the two vectors at this point of contact and using the law of cosines
solved for the resultant vector.
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Forces and vector components ¢  What would the magnitude of the 600N force have to change to so that the resultant of the two forces would align along the posi1ve x-­‐axis? We considered the components of each of the forces along the positive x-axis (the
i component) and the positive y-axis (the j-component). The problem condition
states that the resultant cannot have a j-component so the sum of the jcomponents of the two vectors which are being added together must be equal to
0. Using a sign convention that a positive direction is toward the labeled end of the
axis, we resolve the unknown force (represented by the 600 N vector in the
drawing) into its x and y components and set the magnitude of the y-component
equal to 800 N. This allows us to solve for the magnitude of the unknown vector.
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Cartesian Coordinates ¢  There is a special case when two components that are perpendicular are combined 5
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Cartesian Coordinates ¢  The resultant of these two vectors, F, is formed in the usual manner by connec1ng the tail of the sta1onary vector to the head of the moved vector y
  
F = F1 + F 2
F
F1
x
F2
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Cartesian Coordinates ¢  U1lizing Cartesian coordinates.   
F = F1 + F 2
y
F
Fy
F = F +F
2
x
2
y
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x
Fx
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Cartesian Coordinates ¢  From the drawing, we have the following rela1onships tan (α ) =
Fy
Fx
y
Fy = F sin (α )
Fx = F cos (α )
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Cartesian Coordinates
F
α
Fx
Fy
x
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Resultant of a Series of Vectors ¢  The really nice part of this comes when we take a series of forces at a point and develop a single resultant from all the forces 9
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Resultant of a Series of Vectors ¢  We will start with three forces F1, F2, and F3 and try and find the resultant, F which is the vector sum of the three forces y
F3
F2
x
F1
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Resultant of a Series of Vectors ¢  We have already moved (or the forces were already posi1oned) so that their tails were at the same point y
F3
F2
x
F1
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Resultant of a Series of Vectors ¢  The resultant F is the vector sum of the three forces, F1, F2, and F3    
F = F1 + F 2 + F 3
y
F3
F2
x
F1
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Resultant of a Series of Vectors ¢  Considering the two forces in components. ( )
= F sin ( β )
= F cos (α )
= F sin (α )
y
F1x = F1 cos β
F1y
1
F2x
F2y
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F3
F2
F1x
2
F1y
2
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F2y
α
F2x
β
F1
x
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Resultant of a Series of Vectors ¢  The components of F3 y
F3
F3y
γ
x
F3x
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F4
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Resultant of a Series of Vectors ¢  And subs1tu1ng our trigonometric evalua1ons Fx = F2 cos (α ) − F1 cos ( β ) − F3 cos (γ )
Fy = F2 sin (α ) − F1 sin ( β ) + F3 sin (γ )
y
F3
F3y
γ
x
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F3x
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F4
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Resultant of a Series of Vectors ¢  If we maintain a consistent sign conven1on, we can make a general statement about finding the resultant of any n vectors whose lines of ac1on intersect at a point 16
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Resultant of a Series of Vectors n
Fx = ∑ Fxi
i =1
n
Fy = ∑ Fy i
i =1
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Resultant of a Series of Vectors ¢  Remember that this is only true if we maintain a consistent sign conven1on and take the algebraic sign from our axis selec1on n
Fx = ∑ Fxi
i =1
n
Fy = ∑ Fy i
i =1
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Problem Determine the magnitude
of the resultant force acting
on the corbel and its
direction θ measured
counterclockwise from the
x-axis.
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Problem If the magnitude of the
resultant force acting on
the eyebolt is 600 N and its
direction measured
counterclockwise from the
positive x-axis is θ = 30°,
determine the magnitude of
F1 and the angle ϕ.
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Homework ¢  Remember that you need to have the “Clicker” with you on Wednesday ¢  Problem 2-­‐35 ¢  Problem 2-­‐41 ¢  Problem 2-­‐50 Cartesian Coordinates
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