Equilibrium of Forces: Finding Magnitude and Direction of Unknown Force P

[FaraDazed]

Homework Statement

A particle is in equilbrium with the forces of 4N north, 8N west, 5√2N south east and P.

Find the magnitude and the direction of P

Homework Equations

?

The Attempt at a Solution

A am a bit lost on this question, I can make an educated guess on the magnitude and direction having done what I have done so far...
<br /> y=4-5\sqrt{2}cos45 \\<br /> y=-1 \\<br /> x=-8-5\sqrt{2}sin45 \\<br /> x=-13<br />
This would put the magnitude at near 12N and the direction at about 80° from north. However I am a bit lost on how to actually work it out (and even if what I have done so far is any good).

Any help is appreciated.

 

[tiny-tim]

Hi FaraDazed!

Yes, that looks ok (except that you've got one of the "south-east" signs wrong) …

you're adding the three known vectors, and then your answer will be minus that.

 

[FaraDazed]

Hi FaraDazed!

Yes, that looks ok (except that you've got one of the "south-east" signs wrong) …

you're adding the three known vectors, and then your answer will be minus that.

Hi, thanks for your reply :)

OK, should it be this?
<br /> y=4-5\sqrt{2}cos45 \\<br /> y=-1 \\<br /> x=-8+5\sqrt{2}sin45 \\<br /> x=-3.01<br />

Or this?
<br /> y=4+5\sqrt{2}cos45 \\<br /> y=8.99 \\<br /> x=-8-5\sqrt{2}sin45 \\<br /> x=-13<br />

 

[tiny-tim]

yes!

(but where did the .01 come from?

did you use a calculator for that? )

 

[FaraDazed]

yes!

(but where did the .01 come from?

did you use a calculator for that? )

probably a rounding error, I did 5√2 = 7.07, then did 7.07sin45 = 4.99 (just realized this was wrong as rounding it makes it 5).

Thanks for point that out.

Not sure where to go from there though. Its doing my head in as we have a problem sheet to do and this is one of the easy ones, and I am ok with the harder ones about rough inclined planes with co-efficient of friction etc yet its the 'easy' ones that always get me!

OK well I think I got the magnitude to be 10N as 1^2+3^2=10 , and the direction involves 18.34° because arctan(1/3)=18.34, but not sure if is 90-18.34 = 71.66° from north?

 

[tiny-tim]

Hi FaraDazed!

probably a rounding error, I did 5√2 = 7.07, then did 7.07sin45 = 4.99 (just realized this was wrong as rounding it makes it 5).

hmm … you do know that sin45° is exactly 1/√2, don't you?

Not sure where to go from there though

That gives you the sum of the given forces.

So the balancing force must be minus that sum.

 

[FaraDazed]

Hi FaraDazed!


hmm … you do know that sin45° is exactly 1/√2, don't you?

No, I am not all that up on my surd notation just yet. I am getting better, slowly, though

That gives you the sum of the given forces.

So the balancing force must be minus that sum.

OK cheers, could you check my edit to my last post please?

 

[tiny-tim]

… 5√2N south east …

… could you check my edit to my last post please?

you mean, is it x = -8 - 5 or -8 + 5 ?

what do you think, and why? ​

 

[FaraDazed]

you mean, is it x = -8 - 5 or -8 + 5 ?

what do you think, and why? ​

Its -8+5 because its south east and east on an x-y plane is positive. When doing my original diagram I put the 5√2 as south-west .

I meant could you check the last bit of it actually, I've just c+p it below (and made changes after realsing it is SE not SW) anyway :)

OK well I think I got the magnitude to be 10N as 1^2+3^2=10 , and the direction involves 71.56° because arctan(3/1)=71.56, but as it would be in a south(ish)-west direction it would be 180+71.56=251.56° from north?

 

[tiny-tim]

OK well I think I got the magnitude to be 10N as 1^2+3^2=10

no, it's √10 !

, and the direction involves 71.56° because arctan(3/1)=71.56, but as it would be in a south(ish)-west direction it would be 180+71.56=251.56° from north?

let's see …

P = (3,1), so that's a bit north of east (P is the opposite of the sum of the other forces)

but yes the angle from north is arctan3

 

[FaraDazed]

no, it's √10 !

Of course it is, .

let's see …

P = (3,1), so that's a bit north of east (P is the opposite of the sum of the other forces)

but yes the angle from north is arctan3

Right OK, of course as it needs to balance out, so it would simply be 71.56° from north?

 

[tiny-tim]

Right OK, of course as it needs to balance out, so it would simply be 71.56° from north?

yup!

(though i suspect from the "south-east" in the question that they want you to say "18.44° north of east" )

 

[FaraDazed]

Thanks for your help :)