#1

An object in equilibrium has three forces exerted on it. A(n) 39 N force acts at 69.6 degrees and a 43 N force acts at 32.7 degrees. What is the direction of the third force? (Consider all angles to be measured counterclockwise from the positive x axis) Answer in degrees.

Part B: What is the magnitude of the third force? Answer in units of N.

I pretty much know what I'm doing but there's some of it I don't get.

Fx = 43 sin ____ I don't know what number goes at the end

Fy = 39 - 43 cos ___ again I don't know the number that goes in the underline. I tried it with 32.7 degrees, and I tried it with the difference and they're both wrong. Since I looked at an example and the number they had in the underline wasn't anywhere in the problem.

I also know that to find the degree it's tan(fx/fy). and then that + 180. Is it tan inverse or tan? And then to find the magnitude I just use Pythagorean Theorem.

So all I really need to know is what goes in the fx and fy equations.

Part B: What is the magnitude of the third force? Answer in units of N.

I pretty much know what I'm doing but there's some of it I don't get.

Fx = 43 sin ____ I don't know what number goes at the end

Fy = 39 - 43 cos ___ again I don't know the number that goes in the underline. I tried it with 32.7 degrees, and I tried it with the difference and they're both wrong. Since I looked at an example and the number they had in the underline wasn't anywhere in the problem.

I also know that to find the degree it's tan(fx/fy). and then that + 180. Is it tan inverse or tan? And then to find the magnitude I just use Pythagorean Theorem.

So all I really need to know is what goes in the fx and fy equations.

#2

Asking the pit for pyshics help. . goodluck

#3

3 duh. the answer is always 3 in physics

#4

dude ur so lucky

in physics.....im still learning about gravity half way thro the yr

taught by a 70 yr old man

im in 11th grade.

in physics.....im still learning about gravity half way thro the yr

taught by a 70 yr old man

im in 11th grade.

#5

this is basic stuff you should know if you're taking A levels. I'd usually help, but it's coming up to 1 in the morning and my brain has almost completely shut off.

#6

dude ur so lucky

in physics.....im still learning about gravity half way thro the yr

taught by a 70 yr old man

im in 11th grade.

I wish I was still learning gravity.

My friends and I were going to get together and help each other with it but the snow storm made that not happen. If we have school tomorrow, it's due. I have 3 out of 14 done

#7

Oh god this takes me back...I haven't done physics in years, but I've got an Advanced higher in it...It can't be good that I remember **** all >_<

#8

you need to resolve the vector onto the x and y of the graph. Then you just add them together and then go up on the graph to where they meet to find where the angle is going.

#9

this is basic stuff you should know if you're taking A levels. I'd usually help, but it's coming up to 1 in the morning and my brain has almost completely shut off.

Our teacher isn't that good at teaching. He gives you a few problems in the book as example and then gives you homework based on nothing to really do with the problems that we did. It sucks.

#10

I don't know if your equations are correct. I think they should be

F1x = 39 cos 69.6 N

F1y = 39 sin 69.6 N

F2x = 43 cos 32.7 N

F2y = 43 sin 32.7 N

F3x = F3 cos theta

F3y = F3 sin theta

Fx = F1x + F2x + F3x = 0

Fy = F1y + F2y + F3y = 0

If your angles are measure from the x axis, whenever you're talking about the x coordinate, you should use cosine, and whenever you're talking about the y coordinate, you should use sine.

Plug the F1x, F2x, etc. into the last two equations and solve the system of equations.

F1x = 39 cos 69.6 N

F1y = 39 sin 69.6 N

F2x = 43 cos 32.7 N

F2y = 43 sin 32.7 N

F3x = F3 cos theta

F3y = F3 sin theta

Fx = F1x + F2x + F3x = 0

Fy = F1y + F2y + F3y = 0

If your angles are measure from the x axis, whenever you're talking about the x coordinate, you should use cosine, and whenever you're talking about the y coordinate, you should use sine.

Plug the F1x, F2x, etc. into the last two equations and solve the system of equations.

#11

Fx= 43cos32.7 + 39cos69.6 - Xcos(theta)

Fy= 43sin32.7 + 39sin69.6 - Xsin(theta)

Object is in equilibrium, so Fx=Fy=0

Rearrange those two to find X in each direction.

Xcos(theta) = X in the x direction = Xx

Xsin(theta) = X in the y direction = Xy

theta = arctan(Xy/Xx)

magnitude = (root)(Xx^2+Xy^2)

Edit: darn, beaten by seconds... and by someone who knows how to do subscript letters. I'm very tired and i was typing this out in notepad

Fy= 43sin32.7 + 39sin69.6 - Xsin(theta)

Object is in equilibrium, so Fx=Fy=0

Rearrange those two to find X in each direction.

Xcos(theta) = X in the x direction = Xx

Xsin(theta) = X in the y direction = Xy

theta = arctan(Xy/Xx)

magnitude = (root)(Xx^2+Xy^2)

Edit: darn, beaten by seconds... and by someone who knows how to do subscript letters. I'm very tired and i was typing this out in notepad

*Last edited by James? at Feb 13, 2008,*

#12

Right, this could be slightly sketchy, but I think you need to split each force into two components, their vertical and horizontal parts, then you'd have

[force 1 sin angle] + [force 2 sin angle] + [unknown force sin angle] = [0]

[force 1 cos angle] [force 2 cos angle] + [unknown force cos angle] [0]

Can't quite remember where to go from there, but that's the way I remember doing all these kinda things.

[force 1 sin angle] + [force 2 sin angle] + [unknown force sin angle] = [0]

[force 1 cos angle] [force 2 cos angle] + [unknown force cos angle] [0]

Can't quite remember where to go from there, but that's the way I remember doing all these kinda things.

#13

#14

magnitude = 78N

direction = 230 degrees

there you go

direction = 230 degrees

there you go

#15

Fx = 43 sin 32.7 <last number is the angle>

Fy = 39 cos 69.6 " "

i dunno what that 39 - 43 you wrote is, but this is what it should be. unless you have it completely wrong, it would be great to know the context.

Fy = 39 cos 69.6 " "

i dunno what that 39 - 43 you wrote is, but this is what it should be. unless you have it completely wrong, it would be great to know the context.

#16

I'm guessing the third force is the resultant of the two listed forces added together?

If so, then Fx = 39cos(69.6) + 43 cos(32.7), and Fy = 39cos(69.6) + 43cos(32.7).

The force is then sqrt(Fx^2 + Fy^2), and the direction would be the tan^-1(Fy/Fx).

That is, provided, that your two supplied vectors are added together and not subtracted, or anything else.

If so, then Fx = 39cos(69.6) + 43 cos(32.7), and Fy = 39cos(69.6) + 43cos(32.7).

The force is then sqrt(Fx^2 + Fy^2), and the direction would be the tan^-1(Fy/Fx).

That is, provided, that your two supplied vectors are added together and not subtracted, or anything else.

#17

(clears throat) i can work that out for you in 1 minute but i won't because you won't learn that way. I prefer the sin law or cosine law, instead of the fx,fy method. the sin method is brilliant because its a simulataneous equation that sorts out all three sides in a heartbeat provided ur not a muppet. IT WILL WORK if you do it right despite ur teacher may not have told about it or not recommend it.

if u do use the sin law remember to link up the vectors in the form of a triangle. go to this site http://library.thinkquest.org/20991/alg2/eqtri.html and read the "law of sines" section

its an easy representation of the law.

keep in my mind i love that equation and always use it. So if you use be very careful to link up the vectors from head to tail to form the triangle. (Remember to keep your angles!!!!)

hope that helps.

if u do use the sin law remember to link up the vectors in the form of a triangle. go to this site http://library.thinkquest.org/20991/alg2/eqtri.html and read the "law of sines" section

its an easy representation of the law.

keep in my mind i love that equation and always use it. So if you use be very careful to link up the vectors from head to tail to form the triangle. (Remember to keep your angles!!!!)

hope that helps.

#18

The way I did it (assuming I did it right: it's been a little while since I've done equilibrium) was to convert both given forces to rectangular (x,y) coordinates, add up the parts and then convert back to polar. From there, you'd just have the same magnitude of force on the opposite side of the object (180 + the angle at which the total given force is acting). Here's my equations:

Fx = 39cos(69.6) + 43cos(32.7)

Fy = 39sin(69.6) + 43sin(32.7)

Convert those to polar and you get: (77.8, 50.2)

Add 180 to the angle and you get force three being 77.8N at 230.2 degrees.

I hope I didn't misunderstand the question and draw my diagram wrong though. Yeah, I'm a bit of a physics nerd (I'm a health physics and biology major).

Fx = 39cos(69.6) + 43cos(32.7)

Fy = 39sin(69.6) + 43sin(32.7)

Convert those to polar and you get: (77.8, 50.2)

Add 180 to the angle and you get force three being 77.8N at 230.2 degrees.

I hope I didn't misunderstand the question and draw my diagram wrong though. Yeah, I'm a bit of a physics nerd (I'm a health physics and biology major).

#19

The way I did it (assuming I did it right: it's been a little while since I've done equilibrium) was to convert both given forces to rectangular (x,y) coordinates, add up the parts and then convert back to polar. From there, you'd just have the same magnitude of force on the opposite side of the object (180 + the angle at which the total given force is acting). Here's my equations:

Fx = 39cos(69.6) + 43cos(32.7)

Fy = 39sin(69.6) + 43sin(32.7)

Convert those to polar and you get: (77.8, 50.2)

Add 180 to the angle and you get force three being 77.8N at 230.2 degrees.

I hope I didn't misunderstand the question and draw my diagram wrong though. Yeah, I'm a bit of a physics nerd (I'm a health physics and biology major).

your too late

im an engineering major

#20

your too late

im an engineering major

Ha, yeah, I'm not used to forums moving so quickly (this is my first time posting on any UG forums at all): everyone posted within minutes of one another, lol.

#21

welcome to teh pit

#22

Thanks guys. I worked it out using the fx and fy equations I was given and got the correct answer which you guys got. Thanks.