# Homework Help: This one will tackle your physics skills it tackled mine !

1. Nov 12, 2007

### mrdummy

1. The problem statement, all variables and given/known data

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

2. Relevant equations

F net = F c ( Centripetal force)

3. 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.

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Last edited: Nov 12, 2007
2. Nov 12, 2007

### fliinghier

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

3. Nov 12, 2007

### mrdummy

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

4. Nov 12, 2007

### fliinghier

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

5. Nov 12, 2007

### 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

6. Nov 12, 2007

### fliinghier

7. Nov 12, 2007

### mrdummy

it is a JPG image
i have put a new one
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

8. Nov 12, 2007

### 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.

9. Nov 12, 2007

### 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

10. Nov 12, 2007

### mrdummy

btw how long does it take to approve pictures

11. Nov 12, 2007

### fliinghier

do the strings have mass?

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

12. Nov 12, 2007

### 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

13. Nov 12, 2007

### 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.

14. Nov 12, 2007

### 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 cant get the total force ( tension) in each string as my question asks

15. Nov 12, 2007

### 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?

16. Nov 12, 2007

### mrdummy

is the tension in the upper string
790 Cos 45

17. Nov 12, 2007

### fliinghier

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

18. Nov 12, 2007

### 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 cant get my head around how do i got about finding it

19. Nov 12, 2007

### fliinghier

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