Consider a w-N weight suspended by two wires as shown in the accompanying figure. If the magnitude of the vector F sub 2 is 110 N, find w and the vector F sub 1:

- anonymous

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- anonymous

|dw:1328415051107:dw|

- anonymous

thats what i am trying to find

- anonymous

F2 is 110 N as stated in my problem

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## More answers

- bahrom7893

wait what are u trying to find?

- anonymous

right, but i need to find w=weight of obejct and magnitude of F sub 1

- anonymous

physics?

- anonymous

that is how the problem is written

- anonymous

i believe a system should solve this, with the understanding f sub 1 + F sub 2=w, cant seem too put it together though

- anonymous

how about ratio and proportion to find the value of the magnitude of F sub 1|dw:1328416025545:dw| you can find the value of F sub 1.
and since you said that F sub 1 + F sub 2=w, plug the values to find weight. .
but i'm not sure. .

- anonymous

what answer do you get for the magnitude of Fsub 1

- anonymous

|dw:1328416246651:dw| i don't have calculator. .

- anonymous

i have 124.47 N

- anonymous

the answer was 148.554..

- anonymous

break it into components. Vector(F1)+vector(f2)+vector(w)=0.
Vector f1=magf1<-cos(t1),Sin(t1)>
Vector f2=magf2
X components add to zero
-magf1cos(43)+110cos(38)=0
Solve for magf1 and plug into:
y components:
magf1sin(43)+110sin(38)=w

- anonymous

and that made me wrong. .lol

- anonymous

how do the x components add to zero

- anonymous

instantaneously/or not hah is not in motion. There is no movement in the x direction. Weight is a downward vector so it has no x component. Thus F1 and F2's x components add to zero.

- anonymous

im am still not sure

- anonymous

d|dw:1328417056124:dw|
You have these three vectors to worry about. Adding the x components gives you zero even if you include the weight vector because as you can see in the picture weight has a x-component of zero.

- anonymous

they add to zero because there is no movement.

- anonymous

there is also no movement in the y direction but weight has a -y component.

- anonymous

even if they have different angles, like cos43 and cos 38

- anonymous

those indicate the direction of the vectors. Multiplying them by the magnitude will give you your components.

- anonymous

Notice how I put magf1<-x,y> for vector f1. That vector has a negative x component.

- anonymous

So in my example, based on how you did it, my magnitude for f1 would be given by:|dw:1328417583344:dw|

- anonymous

?

- anonymous

Did you say you have the answers to this problem?

- anonymous

but yes that should be the magnitude of f1 by algebraic methods

- anonymous

its like 148.55...something

- anonymous

for the weight right?

- anonymous

yes for the weight

- anonymous

okay yeah you should be getting 118.52.. for f1

- anonymous

more importantly do you understand?

- anonymous

yes i do thank you very much

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