# Vector Addition Problem - Statics

## Homework Statement

Find the magnitude and direction of the resultant force Fr=F1+F2+F3 by first finding F'=F1+F2 then Fr=F'+F3

known values are in the link

## Homework Equations

Basic vector addition. Law of cosines. Law of sines.

## The Attempt at a Solution

Here's my attempt. I'm not convinced by my answer.
http://imgur.com/LcrFApu

Last edited by a moderator:

Quantum Defect
Homework Helper
Gold Member

## Homework Statement

Find the magnitude and direction of the resultant force Fr=F1+F2+F3 by first finding F'=F1+F2 then Fr=F'+F3

known values are in the link

## Homework Equations

Basic vector addition. Law of cosines. Law of sines.

## The Attempt at a Solution

Here's my attempt. I'm not convinced by my answer.
http://imgur.com/LcrFApu

Another way of calculating the force is using x and y components. Add up all of the x components and all of the y-components of the three vectors to get the x and y components of the final vector. This saves you the trouble of calculating an intermediate vector.

x component = magnitude * cos (theta) -- theta measured from +x axis; y-component = magnitude * sin (theta) -- again theta measured from +x axis.

Last edited by a moderator:
BvU
Science Advisor
Homework Helper
Hello Jack, welcome to PF :)

F' is just fine.
Fr magnitude is fine too. Check the angle calculation. From your drawing you can already see that 29.12 degrees looks to be too low.

Of course, Quantum D gives a good alternative -- that should be equivalent, but I don't know if that's what the exercise wants you to do.

Quantum Defect
Homework Helper
Gold Member
Hello Jack, welcome to PF :)

F' is just fine.
Fr magnitude is fine too. Check the angle calculation. From your drawing you can already see that 29.12 degrees looks to be too low.

Of course, Quantum D gives a good alternative -- that should be equivalent, but I don't know if that's what the exercise wants you to do.

I get essentially the same numbers that are in the box using x, y vectors: 19 degrees, 29.6 N

QD, I would've much rather have done it using component method, however my professor would take off points.

BvU, are you referring to the 29.67N as being the correct magnitude for Fr and 19.12 degrees being incorrect? Could it be 19.12 degrees from the vertical?

I get essentially the same numbers that are in the box using x, y vectors: 19 degrees, 29.6 N
That's reassuring. Thanks for checking!

Quantum Defect
Homework Helper
Gold Member
That's reassuring. Thanks for checking!
My final x components were: +9.7 N x^ direction, +28 N y^ direction -- using inverse tangent gives me 19 degrees from y-axis, as you conclude above.

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BvU
Science Advisor
Homework Helper
To reassure you both: I've been brainwashed to zero degrees ##\equiv## positive x-axis direction. So I find 1.237 (radians, that is. The one and only reasonable unit for angles :) -- did I also say the brainwashing included ##2\pi## for a full circle ?) And I really am a physicist !

So we all agree and Jack can shed his uncertainty on this subject.