addition of vectors phy 115, 201, and 213 important points to remember: sketch the coordinate...
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Addition of VectorsPHY 115, 201, and 213
Important points to remember: Sketch the coordinate system and draw vectors and their components. Label all values including angles.
Pay close attention to signs
Add ‘x’ components together
Add ‘y’ components together
Rectangular form displays a vector’s components
Polar form displays a vector’s magnitude and the angle (CCW from the + ‘x’ axis). Polar form is sometimes referred to as Standard Position. Calculator should be in ‘degree’ mode. When solving right triangles, ‘opposite’ is always the side opposite the angle in question.
The resultant ( R ) is the equivalent of the given vectors. If the given vectors are forces, then R is the equivalent force that could replace the given vectors.
The equilibrant (E) is the vector that offsets (cancels) R. Because R represents the given vectors, then E offsets (cancels) the given vectors. Such a situation is referred to as equilibrium and E is the vector that puts the system in equilibrium.
The Problem:
a) Add these three vectors and find the resultant ( R ):
A = 15 LB @ 25o
B = 35 LB @ 130o
C = 20 LB @ 280o
b) Determine the equilibrant (E) needed to put the system of vectors in equilibrium.
+y
+x
AB
C
A y
A x
C x
C y
B x
B y
40o
A 15 @ 25 14 6.3ox yLB LB
��������������
B 35 @ 130 22 27ox yLB LB
��������������
C 20 @ 280 3.5 20ox yLB LB
��������������
Solve each vector for its components. That is, solve each triangle for the sides. Never mix values when solving a triangle. For example, do not put force units on one side of the triangle and distance units on the other side.
Now, sum like components. Add “x” values to “x” values, etc.
R 4.5 13x yF LB ��������������
These components of the Resultant are the sides of the Resultant triangle.
+y
+x
RR y
R x
R 4.5 13 14 @ 109ox yF LB LB ��������������
Now, solve this triangle for the components of R.
1R 14 @11[10 ] oLB��������������
This vector, R, could replace vectors A, B, and C.
The Equilibrant, E, is 180 o away from R.
+y
+x
RR y
R xE
1E 14 @ 29 [10 ] oLB��������������
1R 14 @11[10 ] oLB��������������
1R 14 @ 11[10 ] 180o oLB ��������������