Adding and subtracting vectors using vector diagrams

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Homework Statement
Classical Mechanics
Relevant Equations
s= (a^2 + b^2 + 2ab cos alpha)^1/2
Printed textbook problem 3: Vector a has magnitude 5.2 units and is directed east. Vector b has magnitude 4.3 units and is directed 35° west of north. Find magnitudes and directions of (a) a + b and (b) a − b by vector diagrams.
Handwritten vector addition solution: diagram of vectors a and b with a 35 degree angle and 55 degree angle, followed by the formula for the magnitude of a plus b with cosine law, giving 4.4587, and tan theta = 1.2885, theta = 52.19 degrees.
 
:welcome:

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

Handwritten equation: magnitude of vector a plus vector b equals 4.4587.


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.