I am sure this is easy but i am stuck

[Vane9488]

The question: A grasshopper makes four jumps. The displacement vectors are (1) 27.0 cm due west (2) 23.0 cm, 35.0^{o} due south of west (3) 28.0 cm, 55.0^{o} south of east and (4) 35.0 cm, 63.0 ^{o} 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 ^{o} 27 cos 0^{o} 27 sin 0^{o}
(2) 23.0 cm 35.0 ^{o} 23 cos 35^{o} 23 sin 35^{o}
(3) 28.0 cm 55.0 ^{o} 28 cos 55^{o} 28 sin 55^{o}
(4) 35.0 cm 63.0 ^{o} 35 cos 63^{o} 35 sin 63^{o}

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=\sqrt{x^{2}+ y^{2}}
r=111.5 cm

the direction i get 38.6^{o}.

Thats not the right answer! Please help!

 

[LowlyPion]

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.

 

[Vane9488]

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!

 

[nmacholl]

I haven't done vector addition in a while but here it goes!

I use D_x 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)

D_2x = -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.)
D_2y = -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.

D_3x = 16.06cm
D_3y = -22.93cm

D_4x = 15.89cm
D_4y = 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 13.03^o S of E

Is that correct?

 

[Vane9488]

I haven't done vector addition in a while but here it goes!

I use D_x 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)

D_2x = -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.)
D_2y = -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.

D_3x = 16.06cm
D_3y = -22.93cm

D_4x = 15.89cm
D_4y = 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 13.03^o S of E

Is that correct?

I don't think so because it asks me to express the direction with respect to due west.

 

[nmacholl]

No because it asks me to express the direction with respect to due west.

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

I hope this helps.

 

[Vane9488]

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

I hope this helps.

Thank You very much but I am still confused

 

[nmacholl]

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:

X_total = D₁ + D_2x + D_3x + D_4x
Y_total = D_2y + D_3y + D_4y

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.

 

[LowlyPion]

I haven't done vector addition in a while but here it goes!

I use D_x 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)

D_2x = -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.)
D_2y = -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.

D_3x = 16.06cm
D_3y = -22.93cm

D_4x = 15.89cm
D_4y = 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 13.03^o 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.

 

[nmacholl]

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 @ 11.64^o S of E or 168.36^o S of W

I think that's better.

 

[LowlyPion]

I didn't run the numbers.

The OP is encouraged to verify to be certain of the proper magnitude.

 

[nvn]

mmacholl: That still doesn't look quite right yet. I currently got R = 14.744 cm at 19.589 deg S of W.