physics help....

like i tried this:

Fg= mg

= (10) (9.81)
Fg= 98.1 N

Sin= opp/hyp

sinO = 1/2 Fg / Ft

Ft= (1/2 Fg) / sinO
Ft= 49.05 / sin(20)
But that isnt right when i checked it against an example

idk why i would multiply by 10
or WTF

this was a step by step example i found, but it doesnt some out right.

Don't worry about the multiply by 10. I think in the problem you're looking at, they have the masses given to you so you can calculate the magnitude of the forces.

there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???

No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.

SEE I WAS RIGHT!!!!! YAY FOR ME!!!!!

I bet you never thought a homeless hobo who smells of booze and cigarettes would discuss physics with you, huh?

we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!

I'm not technically homeless. My local park is my home.

there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???

No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.

SEE I WAS RIGHT!!!!! YAY FOR ME!!!!!

I bet you never thought a homeless hobo who smells of booze and cigarettes would discuss physics with you, huh?

we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!

I'm not technically homeless. My local park is my home.

Can you get to pan handling for spare change so you can buy me a drink! I will be there in a bit!

I already have my booze here. I keep it on me at all times since I'm an alcoholic.

I think it's really an intuative problem. Where is the force going to be pulling based on the angle between the two forces? What quadrant are the forces in? You don't have all the data, so you don't really need to use an equation to figure it out.

If this were a clock, I'm assuming a force at 0 degrees would be pulling toward 12 o'clock. Quadrant 1 is the upper right.

Actually, you can only prove that mathematically if you know the magnitude of the forces. So, my answer would be the third quadrant, Only becaus the two applied forces are in the first quadrant, your reactive force would be in the third.

I agree with this answer if the applied forces are PULLING away from the force balance at 20 and 80 degrees. The reaction force would have to be pulling back at some angle in Q3. I think that's a reasonable assumption to make even though it just says the forces are "applied" at those angles, so really it could be a pulling or pushing force (if pushing at those angles the answer would be Q1). But remembering way back to textbook type questions are written Q3 is the most likely answer.

Even if the forces were flipped in such that they are pushing the ring, the reactive force would sitll be in quadrant 3 since the resultant of the two forces are within quadrant one.

there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???

No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.

SEE I WAS RIGHT!!!!! YAY FOR ME!!!!!

I bet you never thought a homeless hobo who smells of booze and cigarettes would discuss physics with you, huh?

we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!

I'm not technically homeless. My local park is my home.

Can you get to pan handling for spare change so you can buy me a drink! I will be there in a bit!

I already have my booze here. I keep it on me at all times since I'm an alcoholic.

Ok, I am on my way. Don't drink all the booze!

Question:
Two forces are applied to a ring of a force table, one at an angle of 20 degrees, and the other at an angle of 80 degrees. Regardless of the magnitudes of the forces, will the equilibrant be in the
A) first quadrant
second quadrant
C) Third Quadrant
D) Fourth Quadrant
E) You cannot tell which quadrant from the available information....

You're over thinking this. 20 degrees and 80 degrees are both in the first quadrant (0 to 90 degrees). Two vectors in the first quadrant are going to have a sum vector also in the first quadrant.

So the question is: "If you have two forces in the first quadrant, in what quadrant do you need to apply an opposing force to balance the sum of those two?"

The balancing force has to be 180 degrees from the vector sum of the two forces. Since the vector sum is in the first quadrant (somewhere between 0 and 90 degrees), the balancing force has to be in the third quadrant (somewhere between 180 and 270 degrees).

Question:
Two forces are applied to a ring of a force table, one at an angle of 20 degrees, and the other at an angle of 80 degrees. Regardless of the magnitudes of the forces, will the equilibrant be in the
A) first quadrant
second quadrant
C) Third Quadrant
D) Fourth Quadrant
E) You cannot tell which quadrant from the available information....

You're over thinking this. 20 degrees and 80 degrees are both in the first quadrant (0 to 90 degrees). Two vectors in the first quadrant are going to have a sum vector also in the first quadrant.

So the question is: "If you have two forces in the first quadrant, in what quadrant do you need to apply an opposing force to balance the sum of those two?"

The balancing force has to be 180 degrees from the vector sum of the two forces. Since the vector sum is in the first quadrant (somewhere between 0 and 90 degrees), the balancing force has to be in the third quadrant (somewhere between 180 and 270 degrees).

This guy gets it.

seriously...

your name = so faking rad....

i thought it was sofa king rad....


bahahahahahahaha

It's sofa king rad.

Kinda like that band named Sofa Kingdom.

there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???

No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.

But for a force table, the forces pull "out", not "in". Hence, they pull away from the point of intersect.
since both forces are in the 1st quadrant, the resultant equilibrium is also in the 1st quadrant.
The equilibrium is somewhere in between 20 and 80 degrees.

Doesn't matter if they are pushing or pulling. The reactive force to keep it in equilibrium is 180 degrees from the resultant of the two known forces, which puts it in the third quadrant.

Agreed. If the first two forces are pulling, the answer would only be the first quadrant if the third force was for some unexplained reason pushing. And vice versa is the first two are pushing.

I have a BS in Physics and have never heard of a "force table." From the net:

A force board (or force table) is a common physics lab apparatus that has three (or more) chains or cables attached to a center ring. The chains or cables exert forces upon the center ring in three different directions. Typically the experimenter adjusts the direction of the three forces, makes measurements of the amount of force in each direction, and determines the vector sum of three forces. Forces perpendicular to the plane of the force board are typically ignored in the analysis.

Cables or chains can't push. They only pull. So the regardless of the magnitude of any of the forces, the third (equalibriant) force must pull in the third quadrant since the vector sum of the first two pulls in the first quadrant.