I am sure this is easy but i am stuck

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In summary: O = b/a).In summary, the displacement of a grasshopper making four jumps is calculated by adding the x and y components of each jump together. The resulting values, when plugged into the pythagorean theorem and trigonometry equations, give a resultant displacement of 46.97 cm at a direction of 166.97o S of W. It is important to carefully consider the negative signs of each component to accurately determine the magnitude and direction of the resultant displacement.
  • #1
Vane9488
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The question: A grasshopper makes four jumps. The displacement vectors are (1) 27.0 cm due west (2) 23.0 cm, 35.0[tex]^{o}[/tex] due south of west (3) 28.0 cm, 55.0[tex]^{o}[/tex] south of east and (4) 35.0 cm, 63.0 [tex]^{o}[/tex] north of east. Find the magnitude and direction of the resultant displacement. Express the direction with respect to due west.

Vector magnitude direction x component y component
(1) 27.0 cm 0 [tex]^{o}[/tex] 27 cos 0[tex]^{o}[/tex] 27 sin 0[tex]^{o}[/tex]
(2) 23.0 cm 35.0 [tex]^{o}[/tex] 23 cos 35[tex]^{o}[/tex] 23 sin 35[tex]^{o}[/tex]
(3) 28.0 cm 55.0 [tex]^{o}[/tex] 28 cos 55[tex]^{o}[/tex] 28 sin 55[tex]^{o}[/tex]
(4) 35.0 cm 63.0 [tex]^{o}[/tex] 35 cos 63[tex]^{o}[/tex] 35 sin 63[tex]^{o}[/tex]

r= resultant variable

I add all of the x components and I get 62 cm.
I add all of the y components and I get 49.5 cm.

r=[tex]\sqrt{x^{2}+ y^{2}}[/tex]
r=111.5 cm

the direction i get 38.6[tex]^{o}[/tex].

Thats not the right answer! Please help!
 
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  • #2
Welcome to PF.

I'm not sure you have properly accounted for the signs of your magnitudes. For instance with positive x East then your west vector would be negative.
 
  • #3
LowlyPion said:
Welcome to PF.

I'm not sure you have properly accounted for the signs of your magnitudes. For instance with positive x East then your west vector would be negative.


I am still not getting the right answer with this! I am really confused now!
 
  • #4
I haven't done vector addition in a while but here it goes!

I use Dx to label the vectors corresponding to the jumps in sequence. I set up my unit so that west is -x and south is -y which is standard I suppose.

D1 = -27cm (negative because it's going west)

D2x = -13.19cm (you get this by sin(35)*23cm, keep in mind that this is a negative because it's moving westward in the x direction.)
D2y = -18.84cm (Same process of above except cos(35)*23cm)

Now that you got the hang of things I'm going to run down to my solution.

D3x = 16.06cm
D3y = -22.93cm

D4x = 15.89cm
D4y = 31.18cm

Sum of X's = 45.76cm
Sum of Y's = -10.59cm

Make a triangle using those two components and your hypotenuse is the composite vector. 46.97cm 13.03o S of E

Is that correct?
 
  • #5
nmacholl said:
I haven't done vector addition in a while but here it goes!

I use Dx to label the vectors corresponding to the jumps in sequence. I set up my unit so that west is -x and south is -y which is standard I suppose.

D1 = -27cm (negative because it's going west)

D2x = -13.19cm (you get this by sin(35)*23cm, keep in mind that this is a negative because it's moving westward in the x direction.)
D2y = -18.84cm (Same process of above except cos(35)*23cm)

Now that you got the hang of things I'm going to run down to my solution.

D3x = 16.06cm
D3y = -22.93cm

D4x = 15.89cm
D4y = 31.18cm

Sum of X's = 45.76cm
Sum of Y's = -10.59cm

Make a triangle using those two components and your hypotenuse is the composite vector. 46.97cm 13.03o S of E

Is that correct?

I don't think so because it asks me to express the direction with respect to due west.
 
Last edited:
  • #6
Vane9488 said:
No because it asks me to express the direction with respect to due west.

Well that is actually really simple. 13.03o S of E can be converted to W by simple subtraction. 180o - 13.03o = 166.97o
That is to say the new direction is 166.97o S of W.

I hope this helps.
 
  • #7
nmacholl said:
Well that is actually really simple. 13.03o S of E can be converted to W by simple subtraction. 180o - 13.03o = 166.97o
That is to say the new direction is 166.97o S of W.

I hope this helps.

Thank You very much but I am still confused
 
  • #8
Solving vectors for me is basically 3 steps.
1. Setup two equations. One that adds together all the x components and one that adds together all the y components. For you problem my equations looked like this:

Xtotal = D1 + D2x + D3x + D4x
Ytotal = D2y + D3y + D4y

I use the subscripts to number each vector, and the x/y's to label the x or y component. Notice that since D1 is completely due west it is only present in the x equation because it has no y component.

2. Solve for the components. Use trigonometry to draw out each vector and break it into components. Be mindful of the direction of the vector. If it's moving downwards it's y component will have a negative. If it's moving to the right it's x component will have a negative as well.

3. Plug and chug. Plug all your x and y values for D1...D4 into your two equations from step one. This gives you the total displacement in the x and y directions. Using these values (and watching the signs) you can draw a traingle which becomes your composite vector. All you need now are the Pythagorean theorem and an inverse sine or cosine to find the direction.

Hope this clears up my process.
 
  • #9
nmacholl said:
I haven't done vector addition in a while but here it goes!

I use Dx to label the vectors corresponding to the jumps in sequence. I set up my unit so that west is -x and south is -y which is standard I suppose.

D1 = -27cm (negative because it's going west)

D2x = -13.19cm (you get this by sin(35)*23cm, keep in mind that this is a negative because it's moving westward in the x direction.)
D2y = -18.84cm (Same process of above except cos(35)*23cm)

Now that you got the hang of things I'm going to run down to my solution.

D3x = 16.06cm
D3y = -22.93cm

D4x = 15.89cm
D4y = 31.18cm

Sum of X's = 45.76cm
Sum of Y's = -10.59cm

Make a triangle using those two components and your hypotenuse is the composite vector. 46.97cm 13.03o S of E

Is that correct?

Sorry. I think you got your x,y and sin,cos reversed.

Otherwise it looks ok ... except for the result.
 
  • #10
LowlyPion said:
Sorry. I think you got your x,y and sin,cos reversed.

Otherwise it looks ok ... except for the result.

You're right I did flip those.
For my new sums I get 23.79 for x and -4.94 for y.
24.29cm @ 11.64o S of E or 168.36o S of W

I think that's better.
 
  • #11
I didn't run the numbers.

The OP is encouraged to verify to be certain of the proper magnitude.
 
  • #12
mmacholl: That still doesn't look quite right yet. I currently got R = 14.744 cm at 19.589 deg S of W.
 

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