This one will tackle your physics skills it tackled mine

[mrdummy]

Homework Statement

A mass of 5 kg is tied to two strings of equal length, which are attached to a vertical pole at points 2.o m apart. As the pole rotates about its axis, the strings make a right angle with each other. Also, the mass makes 2 revolutions per second.
a) the tension in each string
b) the speed of revolution that makes the lower string go slack

Homework Equations

F net = F c ( Centripetal force)

The Attempt at a Solution

I calculated the total centripal force acting on the mass.
FNET = Fc
Fc = mv^2 / r = m ( 2pi r (revolutions) ) ^2 / R
Fc = 790

I know that the FNETy = 0 because the object is always forming that angle and is constant at it.
Therefore, Force in the y direction on the mass = the total of the forces in the y direction for both the strings and of the mass.
I came up with the equation,
0 = T1 ( top string) SIN 45 - T2 ( bottom string) SIN 45 - MG
0 = T1 ( top string) SIN 45 - T2 ( bottom string) SIN 45 - 10 (5)
0 = T1 ( top string) SIN 45 - T2 ( bottom string) SIN 45 - 50

I do not know how to get the forces in the x direction for the strings.

Please help.

 

[fliinghier]

if the net force in the y direction is zero, which component of the strings' forces provides the centripetal acceleration?

 

[mrdummy]

umm
the only present force is in the x direction
which is 790 N
there is no force in the y direction

 

[fliinghier]

i would suggest drawing free-body diagrams of all the objects involved.

 

[mrdummy]

ok do you want me to draw them and attach them to the image
so you want to see a FBD of the mass

 

[fliinghier]

actually i can't see your image...but it should help you to understand the forces at work.

 

[mrdummy]

it is a JPG image
i have put a new one
which might help you understand better
i forgot to mention in the picture, that the distance betweeen the two strings on the pole, sorry not wall as i wrote on the picture is 2 m

 

[fliinghier]

actually they need to be approved before we can see them. i understand the problem fine, and can draw the picture as well. I'm just asking if you have labeled all of the forces at work on the strings and the mass.

 

[mrdummy]

the forces on the strings
- gravity
- the force of the object
- maybe the reaction force from the pole

the forces on the mass
- gravity
- the two strings pulling it

 

[mrdummy]

btw how long does it take to approve pictures

 

[fliinghier]

do the strings have mass?

also, i have no clue, but you could upload them to photobucket

 

[mrdummy]

um no consider the strings have no mass
i haven't done physics with the mass of the strings involved too
photo bucket?
what is that

 

[fliinghier]

photobucket is a site you can upload pics to in order to link them, but yours are approved so don't bother.

you seem to have the right idea in your drawings. if you look at the forces on the strings, you see that they are the tension which causes the Acent for the object, and the tension which keeps the object from falling due to gravity. therefore the total force on both strings needs to nullify these two other forces.

 

[mrdummy]

i kind of got what you meant
but can you help me understand
how to get the force in the x direction for each string
so i can't get the total force ( tension) in each string as my question asks

 

[fliinghier]

here's a hint: what should the tension in the X and Y directions on each string be, since the angle is 45 degrees?

 

[mrdummy]

is the tension in the upper string
790 Cos 45

 

[fliinghier]

if that were the answer, then the tension in each string would be equal. I'm fairly certain this is not the case.

 

[mrdummy]

yah you are right
i am blank
i understand that the tension in each string is the hypotnuse of a 90-45-45 triange
but i still can't get my head around how do i got about finding it

 

[fliinghier]

well the y components of the tension are in opposite directions and yet add up to Fg

what about the x components?

 

[mrdummy]

no no
look at the picture
the y component of tension in the upper string ( US Y ) equals
the y component of tension in the lower sting ( LS Y ) and the force pulling the weight down which is mg which is 50 N
US Y = LS Y + 50 N

 

[fliinghier]

if you give each of these signs describing up/down, you have:

Fg+US Y+LS Y=0

the LS is negative, as is Fg, while US is positive. US plus LS are equal and opposite to Fg

 

[mrdummy]

yes u r right
US - LS = Fg

 

[mrdummy]

but how do i find y tension is upper string
and y tension in lower string

 

[fliinghier]

what does US X+LS X equal?

 

[mrdummy]

dont they equal each other
they both r in the same direction

 

[mrdummy]

its the y component of tension that will cause a change in tension in each of the strings

 

[fliinghier]

they are the same direction, but why are they equal?

what is equal?

 

[mrdummy]

why wouldn't the tension in the x direction for both the strings be the same..
they both r connected to the same pole

 

[mrdummy]

please can you give me the answer
i m starting to get annoyed by the question

 

[fliinghier]

that doesn't mean they are the same. the mass pulls more on the top string because the mass is below it, so the top string fights gravity and the lower string in the y direction. the x and y of both strings are ___?