Adding and subtracting vectors using vector diagrams

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SUMMARY

The discussion focuses on the correct methods for adding and subtracting vectors using vector diagrams. Key mistakes identified include the incorrect drawing of vectors and the misuse of the cosine law formula. The correct formula for the cosine law is established as c^2 = a^2 + b^2 - 2ab cos α, where α is the angle between the vectors. The correct magnitude for the sum of vectors |\vec a + \vec b| is confirmed to be approximately 4.4587, while the magnitude for the difference |\vec S| should be around 8.4 if the diagram were drawn correctly.

PREREQUISITES
  • Understanding of vector addition and subtraction
  • Familiarity with the cosine law in trigonometry
  • Ability to interpret vector diagrams
  • Knowledge of radians and degrees in angle measurement
NEXT STEPS
  • Study the construction of vector diagrams for accurate representation
  • Learn the application of the cosine law in vector calculations
  • Explore the differences between vector addition and subtraction geometrically
  • Review trigonometric functions in both radians and degrees
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Students and professionals in physics, engineering, and mathematics who are involved in vector analysis and require a clear understanding of vector operations and diagrammatic representations.

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Homework Statement
Classical Mechanics
Relevant Equations
s= (a^2 + b^2 + 2ab cos alpha)^1/2
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:welcome:

Look at the picture:
Doesn't it seem strange that

1758800807980.webp


Check your math -- and is your calculator set up for radians or for degrees ?

##\ ##
 
Ishfa said:
Relevant Equations: s= (a^2 + b^2 + 2ab cos alpha)^1/2

View attachment 365813View attachment 365814
It would be best to sort out part a) first. There are some mistakes. The first two are:

1) The diagram is wrong. When adding vectors using a diagram, you draw the vectors 'tip-to-tail' or construct a suitable parallelogram.

2) The correct formula for the cosine law is ##c^2 = a^2 + b^2~–~ 2ab \cos\alpha## (where ##\alpha## is the internal angle between sides a and b). But you have used a plus sign.

Surprisingly I agree with ##|\vec a + \vec b| = 4.4587##. It looks like some mistakes cancelled!

Minor edit.
 
Last edited:
Steve4Physics said:
Surprisingly I agree with ##|\vec a + \vec b| = 4.4587##. It looks like some mistakes cancelled!
The expressions for the magnitudes are obtained from
##|\mathbf a + \mathbf b|^2=(\mathbf a + \mathbf b)\cdot(\mathbf a + \mathbf b)=a^2+b^2+2ab\cos\alpha##
##|\mathbf a - \mathbf b|^2=(\mathbf a - \mathbf b)\cdot(\mathbf a - \mathbf b)=a^2+b^2-2ab\cos\alpha##
where ##\alpha## is the angle between the two vectors when placed tail-to-tail, here 125°.

In the vector diagram, ##\vec S## is the difference, but the calculation below it is correct for the sum.
 
kuruman said:
In the vector diagram, ##\vec S## is the difference, but the calculation below it is correct for the sum.
The OP has incorrectly drawn the Post #1 diagram, believing (wrongly) that ##\vec s = \vec a + \vec b##.

Then they have ignored their diagram and used:
Ishfa said:
Relevant Equations: s= (a^2 + b^2 + 2ab cos alpha)^1/2
which gives the correct value.

The OP should note thast the question specifically says “By constructing vector diagrams, find the magnitudes and directions of …”.

If the OP had used their (incorrect) diagram correctly, they would have obtained ##|\vec S| \approx 8.4##.
 
Steve4Physics said:
The OP has incorrectly drawn the Post #1 diagram, believing (wrongly) that ##\vec s = \vec a + \vec b##.

Then they have ignored their diagram and used:

which gives the correct value.

The OP should note thast the question specifically says “By constructing vector diagrams, find the magnitudes and directions of …”.

If the OP had used their (incorrect) diagram correctly, they would have obtained ##|\vec S| \approx 8.4##.
That's another way of looking at it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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