Graphical method to find magnitude and direction

[xupe33jrm]

Homework Statement

The force vector FA has a magnitude of 90.0 Newtons and points
due east. The force vector FB has a magnitude of 135 Newtons and
points 75° north of east. Use the graphical method and find the
magnitude and direction of (a) FA - FB (give the direction with respect
to due east) and (b) FB + FA (give the direction with respect
to due west).

The Attempt at a Solution

So confused here. I have tried this all kinds of ways and cannot get the answer the book came up with.
I have tried:
a. R = sqrt((90)^2 + (135)^2) = 162 N
For direction I did tan-1 (135/90) = 56.3 degrees South of east.
B. Same answer just N. of east.

The book sais the answer are 142 N for both and 67 degrees.
Help! I need step by step because I just don't get it. This is not for a homework assignment I am just doing the odds out of the back of the book for practice for our test.

 

[magwas]

It is talking about a graphical method. What about drawing the vectors, adding/substracting them and measure the result. You add two vectors by starting the second at the end of the first, and you substract two vectors by ending the second at the end of the first.

For a coordinate geometry solution you should express x and y coordinates of the vectors, and add/substract them.

You need definition of sine and cosine to convert from (length,angle) to (x,y). You need pythagoras and definition of tangent to convert back.

 

[tiny-tim]

Welcome to PF!

Hi xupe33jrm! Welcome to PF!

(have a square-root: √ and try using the X² tag just above the Reply box )

…I have tried:
a. R = sqrt((90)^2 + (135)^2) = 162 N
For direction I did tan-1 (135/90) = 56.3 degrees South of east.

No, that would be if the 90 and 135 were perpendicular.

Draw a vector triangle …

two of the sides will be 90 and 135 …

now use the cosine and/or sine formula.

What do you get? ​

 

[xupe33jrm]

Could someone maybe draw me a diagram on how to set it up, because I am just not understanding. This problem is driving me nuts. If I knew how to draw it maybe I can figure out the math! thanks!

 

[tiny-tim]

Could someone maybe draw me a diagram on how to set it up, because I am just not understanding. This problem is driving me nuts. If I knew how to draw it maybe I can figure out the math! thanks!

Hi xupe33jrm!

FA: 135 "horizontal", arrow pointing right.

FB: 90 slanting up and to the right, almost "vertical", arrow slanting up and to the right.

FA - FB: Start both FA and FB at the same place, then FA - FB is the unknown length joining the ends, arrow pointing down (so that you follow the arrows: FB + (FA - FB) = FA ).

FA + FB: Start FB from the end of FA (not the beginning), then FA + FB is the unknown length joining the ends, arrow pointing up (so that you follow the arrows: FA + FB = (FA + FB))

 

[xupe33jrm]

Why would 90 not be on the east horizontal since it points due east? Also when trying to find the answer what equation do I use. I am really confused and I know it is probably a simple problem.
Is the drawing I included look right and if so am I trying to find the green side?

 

[tiny-tim]

Why would 90 not be on the east horizontal since it points due east?

oh sorry, i got the 90 and the 135 the wrong way round.

Also when trying to find the answer what equation do I use. I am really confused and I know it is probably a simple problem.
Is the drawing I included look right and if so am I trying to find the green side?

Yes, but you should always put the arrows in, or you may end up drawing the wrong triangle.

Now use the cosine formula.

 

[xupe33jrm]

I got the answer, thanks so much guys. I was getting very angry with this problem, but know it seems so easy!

 

[camrik55]

I still don't get this problem can someone explain please!

 

[tiny-tim]

welcome to pf!

hi camrik55! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[camrik55]

I finally found it out. Thank you though. I am sure I will have more questions very soon!

 

[camrik55]

I've been trying to figure this out for hours and I know it is easy but it keeps tricking me up. I know you need to use the equation W=(F cos theta) but my answer doesn't seem to come out. Here is my problem

A force does +170 J of work when it acts on a moving object and its direction is in the same direction as the object’s displacement. How much work does this force do when the angle between it and the object’s displacement is 56°?

 

[tiny-tim]

hi camrik55!

(have a theta: θ )

A force does +170 J of work when it acts on a moving object and its direction is in the same direction as the object’s displacement. How much work does this force do when the angle between it and the object’s displacement is 56°?

the force, F, and the https://www.physicsforums.com/library.php?do=view_item&itemid=378", d, are the same in both cases …

so write out the equation for F d and 170, and then see how it changes if you change the angle to 56° (from 0°)

 

[isyndhrea]

i don't get this problem.
what is the formula for this problem?

 

[cepheid]

The relevant equation is W = Fdcos(theta) as someone already mentioned.

 

[isyndhrea]

based to my friend the graph is like this..?

 

[Xisune]

My mistake, yes the graph is correct.

 

[tiny-tim]

The force vector FA has a magnitude of 90.0 Newtons and points
due east. The force vector FB has a magnitude of 135 Newtons and
points 75° north of east. Use the graphical method and find the
magnitude and direction of (a) FA - FB (give the direction with respect
to due east) and (b) FB + FA (give the direction with respect
to due west).

based to my friend the graph is like this..?

for F_A + F_B, yes

 

[isyndhrea]

what graph is correct?

is this problem use pythagorean theorem or component method?

 

[tiny-tim]

you can always use either the cosine formula or the component method

(pythagoras only works for right-angled triangles … for general triangles you need the cosine formula)

 

[isyndhrea]

actually this is in my book .. FB-FA due to west

and first i used component method which is

x=90 y=0
x= 35 y=130

sum= x=125 y= 130

then i used pythgorean theorem

 

[isyndhrea]

. cosine formula?.. our professor did not teach about that..

 

[tiny-tim]

actually this is in my book .. FB-FA due to west

and first i used component method which is

x=90 y=0
x= 35 y=130

sum= x=125 y= 130

then i used pythgorean theorem

(that's FB plus FA)

ah, i see what you mean …

yes, when you have the x and y components, you can then use the pythgorean theorem to give you the magnitude

(your answer should be "a magnitude of … Newtons at an angle of … north of west")

 

[isyndhrea]

yes, when you have the x and y components, you can then use the pythgorean theorem to give you the magnitude

(your answer should be "a magnitude of … Newtons at an angle of … north of west")[/QUOTE]


-is only one answer for this problem?..

 

[isyndhrea]

but the result is not 142 ang 67 degree.

 

[isyndhrea]

can you give me a specific formula?

 

[tiny-tim]

that's 67° north of east, the question asks for the angle with respect to west

 

[tiny-tim]

actually this is in my book .. FB-FA due to west

and first i used component method which is

x=90 y=0
x= 35 y=130

sum= x=125 y= 130

then i used pythgorean theorem

also, you've added the components instead of subtracting them

 

[isyndhrea]

so if i subtract the angle will be in north of west

 

[alex.hs]

Angle Between vector and axis

Hi guys...
I am new to his website and i don't know if i am using the correct place to ask my question:shy:
Well the question is:
When a vector is given say A: a i^+b j^+ck^ then how to find the angle between the vector and x-axis: y- axis; z-axis
Thanq for ur time ...