Physics is the bane of my existence
 

Hey guys I can't figure out this problem for the life of me....

A 3.50 kg object is attached to a vertical rod by two strings as in the figure below. The object rotates in a horizontal circle at constant speed 4.95 m/s.

http://i274.photobucket.com/albums/j...sh/physics.gif

What is the tension on the lower and upper string D:?

You come to us when you have a school problem.. hmph!

O.o would you like to me to come to you other times too o.O? I quietly help out around here .-. /gives lots of money when asked for FF events D;

Aw ok I'm sorry.. I'll take that back.. kekeke

lol....no one in my class can get it either D; i think this problem is cursed

Do you have a formula for tension? That's the formula for force right?

I wonder if Lam knows. Kekeke. He just posted a super long thing on Physics. If I see him online, I'll have him look at this.


/has never taken a Physics class in my life

F=MA...tension is essentially force.

It should be like this I think

cos θ = 1.5/ 2
= 41.41 degrees
T1y = mg
T1 sin θ = mg
T1 sin 41.41 = 3.5(9.8)
T1 = 51.85N

but it doesn't work D;

T1 = 49.28 N?

Bleh.

Try centripetal motion equation.

This was good practice for my upcoming physics exam too... so here we go.

You have two identical strings, which are 2.00m each.

First, find the angle the upper string (T1) makes with the upper half of the pole, which is:

cos θ = 1.50 / 2.00
θ = 41.41°

This also applies to the lower half, which will be referred to as (T2).

T1 and T2 combined result in the circular motion of the ball. The horizontal components are what causes the ball to move, so:

T1sinθ + T2sinθ = (mv^2)/r

The radius is part of the horizontal circular motion as well, so you must take the horizontal component. s = string

r = s(sinθ)
r = 2.00sin(41.41)
r = 1.32

Therefore:

T1sinθ + T2sinθ = (mv^2)/r
T1sin41.41 + T2sin41.41 = (3.50 * 4.95^2)/1.32
T1 + T2 = 65.0/(sin41.41)
T1 + T2 = 98.3 N

Now, analyze the vertical components. However, gravity also acts upon the object, so you must take that into consideration with the lower tension.

Components:

T1cosθ
T2cosθ
mg

T1cosθ = mg + T2cosθ
T1cos41.41 = (3.5 * 9.8) + T2cos41.41
T1 - T2 = 34.3 / (cos41.41)
T1 - T2 = 45.7 N

Thus:

T1 + T2 + (T1 - T2) = 98.3 + 45.7
2T1 = 144
T1 = 72 N

Plug the newly found T1 into T1 + T2 to get:

T1 + T2 = 98.3
72 + T2 = 98.3
T2 = 26.3 N

Answers:
T1 = 72.0 N
T2 = 26.3 N

You are such a geek. lol

*dies from post #11*

42.

MIDTERM IN A WEEK!!

AHHH!

It actually pertains to some of the stuff we did earlier... lulz.

I concur.....WTH? LOLZ....math was my WORST

Okay that makes sense. -high fives you- Thanks a lot ^^! Teacher was like....erm you actually got this? I told her one of my friends helped me xP

My teacher would LOVE you

Well...Fuck...