two vector having equal magnitude A makes an angle "theta" with each other. find the magnitude and direction of the resultant.
Asked by riya sinha
| 15th Jun, 2013,
08:01: AM
Expert Answer:
Pick the x axis as half way between the two. Call them A and B, with A below the x axis by ?/2 and B above by ?/2
Use normal vector arithmetic, get the x and y components of each
Ax = Acos(?/2)
Ay = Asin(?/2)
Bx = Acos(?/2)
By = Asin(?/2)
add the components to get the components of the resultant
Rx = 2Acos(?/2)
Ry = 0
combine the components to get magnitude and angle.
with one zero, it is just the other component.
R = ?(Rx² + Ry²)
R = 2Acos(?/2)
angle = tan^-1[ (Ry/Rx) ]
With Ry=0, that becomes 0
Angle = 0
Angle will be halfway between the two.
Use normal vector arithmetic, get the x and y components of each
Ax = Acos(?/2)
Ay = Asin(?/2)
Bx = Acos(?/2)
By = Asin(?/2)
add the components to get the components of the resultant
Rx = 2Acos(?/2)
Ry = 0
combine the components to get magnitude and angle.
with one zero, it is just the other component.
R = ?(Rx² + Ry²)
R = 2Acos(?/2)
angle = tan^-1[ (Ry/Rx) ]
With Ry=0, that becomes 0
Angle = 0
Angle will be halfway between the two.
Answered by
| 15th Jun, 2013,
05:13: PM
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