Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A = A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N R₂ = Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₁ = A cos a = B₂ =B cos b = R₂-A₂+ B₂ = A = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B, = B cos b = R₂-A₂+B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R): Quadrant R = √ (R₂²+R₂²) = Direction:q tan [R, /R]

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Compare the magnitudes of the equilibrant vectors measured from the experiment with
those obtained from the graphical and component methods.
Example: A: 200 g 60° above +x axis
B: 300 g 45° above -x axis
C: 400 g 30°below -x-axis
A, A cos a = 1.96 N x cos 60° = 0.98 N
B₂B cos b = 2.94 N x cos 135º = -2.08 N
C, C cos g = 3.92 N x cos 210°= -3.39 N
R₂-A, + B + C₂ = -4.49 N
A, A sin a = 1.96 N x sin 60° = 1.70 N
By
B sin b = 2.94 N x sin 135º = 2.08 N
C, C sin g = 3.92 N x sin 210° = -1.96 N
Ry= Ay+ By + Cy = 1.82 N
Questions:
I: (a) A: 200 g along +x axis
B: 100 g 45° above -x axis
A₂ = A cos a =
B₂ =B cos b =
R₂-A₂+ B₁₂=
A₂ = A sin a =
B, = B sin b =
R₂ = A + B₂ =
R=(R₂. R₂):_ Quadrant
R = √ (R₂²+R₂²) =
Direction:q=tan [R, /R.]
(c)
A: 100 g along -y axis
B: 200 g along -x axis
A = A cos a =
B = B cos b =
R₂-A₂+ B₂ =
A = A sin a =
B, B sin b =
R=A, +B₂ =
R=(R₁, R₂): Quadrant
R = √ (R₂²+R₂²) =
Direction:q tan¹ [R, /R]
Transcribed Image Text:Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]
II: (a) A: 150 g 60° along +y axis
B: 200 g 45° above -x axis
C: 100 g 30° below -x axis
A₂ = A cos a =
B, =B cos b =
C₁ = C cos g =
R₂-A₂+B+C₂
A, = A sin a =
B, =B sin b =
C₂ = C sin g =
R₂ = A + B + C =
R=(R₁, R₂):
R= √ (R₂²+R²) =
Quadrant
Direction:q=tan [R, R.]
1
(c) A: 200 g along +y axis
B: 100 g 60° above +x axis
C: 200 g 45° below +x axis
A₁ = A cos a =
B=B cos b =
C₁=C cos g =
R₂-A₂+B+C₂
A₁ = A sin a =
B, =B sin b =
C₁ = C sin g =
R= A + B + C₂ =
R=(R₁, R₂): _
R=√(R₂²+R₂²) =
Direction:q=tan¹ [R₂/R₂] =
Quadrant
Transcribed Image Text:II: (a) A: 150 g 60° along +y axis B: 200 g 45° above -x axis C: 100 g 30° below -x axis A₂ = A cos a = B, =B cos b = C₁ = C cos g = R₂-A₂+B+C₂ A, = A sin a = B, =B sin b = C₂ = C sin g = R₂ = A + B + C = R=(R₁, R₂): R= √ (R₂²+R²) = Quadrant Direction:q=tan [R, R.] 1 (c) A: 200 g along +y axis B: 100 g 60° above +x axis C: 200 g 45° below +x axis A₁ = A cos a = B=B cos b = C₁=C cos g = R₂-A₂+B+C₂ A₁ = A sin a = B, =B sin b = C₁ = C sin g = R= A + B + C₂ = R=(R₁, R₂): _ R=√(R₂²+R₂²) = Direction:q=tan¹ [R₂/R₂] = Quadrant
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