Resolving Vector Components

**unstopabl3** · October 18th 2009, 06:09 AM

Hi, I need help with resolving components of vector forces.

For example how would you solve the last question mentioned on this website

Code:

http://www.s-cool.co.uk/alevel/physics/vectors-and-scalars-and-linear-motion/resolving-vectors-into-components.html

I've read the procedure on the above page but still confused on how to solve such questions.

Besides the above question I'm having difficulty with the following:

Firstly I can't figure out how to draw the components.

Secondly, once I have drawn the components I can't figure out which component/force should be Fsintheta and which one should be Fcostheta.

Please provide details on how to do the above.

Thanks!

**skeeter** · October 18th 2009, 06:48 AM

S-Cool | Resolving Vectors into Components

OP wants help w/ the last question.

**unstopabl3** · October 18th 2009, 06:54 AM

huh?

**skeeter** · October 18th 2009, 07:14 AM

Originally Posted by unstopabl3

huh?

I made your reference into a link that is "clickable" ...

**unstopabl3** · October 18th 2009, 07:21 AM

I thought direct links are not allowed thus I coded mine. Thanks though, I just need someone to help me now

**e^(i*pi)** · October 18th 2009, 07:23 AM

Originally Posted by unstopabl3

Hi, I need help with resolving components of vector forces.

For example how would you solve the last question mentioned on this website

Code:

http://www.s-cool.co.uk/alevel/physics/vectors-and-scalars-and-linear-motion/resolving-vectors-into-components.html

I've read the procedure on the above page but still confused on how to solve such questions.

Besides the above question I'm having difficulty with the following:

Firstly I can't figure out how to draw the components.

Secondly, once I have drawn the components I can't figure out which component/force should be Fsintheta and which one should be Fcostheta.

Please provide details on how to do the above.

Thanks!

It doesn't matter in this example because $cos(45) = sin(45) = \frac{\sqrt2}{2}$

**unstopabl3** · October 18th 2009, 07:33 AM

Yea I figured, but how would you go on about solving it?

Also can you answer my other questions.

Thanks.

**e^(i*pi)** · October 18th 2009, 07:42 AM

Originally Posted by unstopabl3

Yea I figured, but how would you go on about solving it?

Also can you answer my other questions.

Thanks.

I resolved the force of the wind to find it's vertical component which I get to be $F_v = -300sin(45) = -150\sqrt2$

This would give a resultant force of $1000-(500+150\sqrt2)$ but the site says that is wrong and I've no idea how they got their answer

**unstopabl3** · October 18th 2009, 07:49 AM

Thanks, could you tell me step by step on how to resolve components in such questions?

Thanks.

**skeeter** · October 18th 2009, 08:22 AM

resultant's x-component, $R_x = 300\cos(45)$

resultant's y-component, $R_y = 500-1000-300\sin(45)$

$|R| = \sqrt{R_x^2 + R_y^2} \approx 357.6$ Newtons

**unstopabl3** · October 18th 2009, 08:42 AM

Could you please explain to me (with diagrams if possible) as how you got those components for that question.

Thanks!

**mathaddict** · October 18th 2009, 08:51 AM

Originally Posted by skeeter

resultant's x-component, $R_x = 300\cos(45)$

resultant's y-component, $R_y = 500-1000-300\sin(45)$

$|R| = \sqrt{R_x^2 + R_y^2} \approx 357.6$ Newtons

HI

shouldn't Rx be 300 sin 45 ?

and the component in Ry = 300 cos 45 ?

**mathaddict** · October 18th 2009, 08:55 AM

Originally Posted by unstopabl3

Could you please explain to me (with diagrams if possible) as how you got those components for that question.

Thanks!

HI

you can draw a triangle with 300 being the hypothenus and 45 degree being the angle on top . Its just exactly the same in the animation and you just add an extra line at the bottom .

Then use some basic trigo to get them .

**unstopabl3** · October 18th 2009, 09:01 AM

That's how did it and got the answer right just wanted to make sure because I got confused with "skeeter's" last post, where I thought he may have made a mistake with the X and Y components. I guess you think same as well. The answer came same either ways because the angle value is same for cos and sin

Besides the above question I'm having difficulty with the following:

Firstly I can't figure out how to draw the components.

Secondly, once I have drawn the components I can't figure out which component/force should be Fsintheta and which one should be Fcostheta.

Please provide details on how to do the above.

Thanks.

**mathaddict** · October 18th 2009, 08:05 PM

Originally Posted by unstopabl3

That's how did it and got the answer right just wanted to make sure because I got confused with "skeeter's" last post, where I thought he may have made a mistake with the X and Y components. I guess you think same as well. The answer came same either ways because the angle value is same for cos and sin

Besides the above question I'm having difficulty with the following:

Firstly I can't figure out how to draw the components.

Secondly, once I have drawn the components I can't figure out which component/force should be Fsintheta and which one should be Fcostheta.

Please provide details on how to do the above.

Thanks.

Vector Resolution

Try reading that . It might help .

Or see the diagram i attached . Sorry for the unclear image though .

You will see a triangle with F being the hypothenus , Fx and Fy are respectively 2 sides of the triangle .

So if you want resolve horizontally, Fx , you need a trigonometric ratio to relate Fx and F . So in this case , you will use

cos theta = Fx / F

so Fx = F cos theta .

Same goes to resolving vertically , find a trigonometric ratio which would relate Fy and F .

So would it be sin ?

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