# Mechanics tension in string

1. Mar 18, 2010

### iloveannaw

This isn't really homework, just doing some revision.

1. The problem statement, all variables and given/known data
Find the tension in each string. Two spheres both of mass, m = 15kg. Diameter of each is 25cm. Each string 35cm long. Spheres are suspended on strings with other end attached to same point on ceiling. Therefore spheres are touching. Hopefully, that is not too difficult to understand, sorry no picture!

Taking g as 10 m/s/s

2. Relevant equations plus attempt
A vertical line drawn from point where string attached to ceiling makes a triangle with string + radius (hypotenuse) and radius of ball (opposite side).
hypotenuse = 0.35 + 0.125 = 0.475 meters
therefore angle alpha = arcsin(opp/ hyp) = arcsin(0.125 / 0.475)

Tension in each secondary string = T2 = mgcos(alpha) = 145 Newtons

However answer at back of book gives 152N - how can this be? g cannot be greater than 10! Really would like some help

2. Mar 18, 2010

### kuruman

How many forces act on each sphere?

3. Mar 19, 2010

### iloveannaw

well, i'd say for each sphere gravity (mg), which pulls directly downward, causing a tension in the string, which acts along string at an angle alpha from the vertical. also the spheres are in contact, so there is a contact force in the x-direction.

still i dont understand. if g = 10 m/s/s, then weight of each sphere is 150N. how can the tension in each supporting string be greater than weight of the thing it supports. I would love an example.

4. Mar 19, 2010

### kuruman

You have an example. Draw a free body diagram for one of the spheres and put in all the forces. In what direction is the contact force? Say that the sum of all vertical components is zero, solve for the tension and see what you get.

5. Mar 19, 2010

### iloveannaw

thank you!

sum Fy = 0 = Tsin(a) - mg >>> Tsin(a) = mg

sum Fx = mgsin(a)cos(a)

T = sqrt(Fx^2 + Fy^2) = 152 , when g = 9.8

is that what you were hinting at?

6. Mar 19, 2010

### kuruman

Not exactly. Look at your first equation. It says

Tsin(a) = mg

Can you solve it to find T?
Given that the angle is not zero, is T greater than, equal to or less than mg?

7. Mar 20, 2010

### iloveannaw

T >mg but I've worked the angle, a, to be 15.3 deg.

T = mg/ sin(a) = 15*10 / sin(15.3) = 568N

this is far too large.
if length of string = 0.35m and radius of sphere = 0.125m, we have a triangle of:
opposite side = radius = 0.125m
hypotenuse = string + radius = 0.475m
therefore sin(a) = opp/hyp = 0.125/ 0.475
and angle a = arcsin(0.125/ 0.475) = 15.3 deg

but i do see how T can be greater than mg now (at least mathematically), otherwise it still seems like a rather weird idea!