Vector Problems: Resultant Magnitude and Cross Product Calculations

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SUMMARY

This discussion focuses on solving vector problems involving resultant magnitudes and cross product calculations. The first problem involves two forces of equal magnitude, where reversing one vector halves the resultant, leading to the conclusion that tan(θ) can be derived from the relationship between the forces. The second problem states that the resultant of two vectors with magnitudes 5 and 4 is 1, indicating that the cosine of the angle between them is -1, which implies they are in opposite directions. These calculations are essential for understanding vector dynamics in physics.

PREREQUISITES
  • Understanding of vector addition and subtraction
  • Familiarity with trigonometric functions, specifically tangent and cosine
  • Knowledge of the cross product and its geometric interpretation
  • Basic skills in setting up coordinate systems for vector analysis
NEXT STEPS
  • Study vector addition and the parallelogram law in detail
  • Learn about the properties of the cross product in three-dimensional space
  • Explore the implications of vector angles on resultant magnitudes
  • Practice solving problems involving forces and their resultant vectors
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone involved in mechanics or engineering who needs to understand vector operations and their applications in real-world scenarios.

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Homework Statement



1. Two forces of equal magnitude are acting on a particle along two different directions. Let [tex]\theta[/tex]
be the angle between them. If direction of one vector is reversed , the magnitude of resultant is halved. Find tan[tex]\theta[/tex].

2.The resultant of the two vectors having magnitudes 5 and 4 is 1. what is the magnitude of their cross product?

Homework Equations





The Attempt at a Solution


for 1st one-
tan[tex]\theta[/tex] = Acos[tex]\theta[/tex][tex]\frac{}{}[/tex]A + Asin[tex]\theta[/tex]
and tan2[tex]\theta[/tex] = 2Acos[tex]\theta[/tex][tex]\frac{}{}[/tex]A + Asin[tex]\theta[/tex]
2nd one -
i get cos[tex]\theta[/tex] as -1.
 
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i can't really work out what you have attempted, note that you can put a whole equation in the tex tags.

first if they are parallel, the resultant would go to zero when one is reversed
second, if they are pepindicular, the resultant would not change.

so you know the original angle is some where between 0 & pi/2

i would start by setting up a coordinate axis, where x is parallel to the vector that will be unchanged, then
[tex]F_x = F + F cos(\theta)[/tex]
[tex]F_y = F sin(\theta)[/tex]
 

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