Finding Tension in Suspended Spheres with Given Mass and String Length

In summary, the tension in each supporting string is greater than the weight of the spheres it supports.
  • #1
iloveannaw
45
0
This isn't really homework, just doing some revision.

Homework Statement


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. Homework 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
 
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  • #2
How many forces act on each sphere?
 
  • #3
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 don't 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
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
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
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
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!
 

Related to Finding Tension in Suspended Spheres with Given Mass and String Length

What is tension in a string?

Tension is the force exerted by a string when it is pulled tight. It is measured in units of newtons (N).

How is tension in a string calculated?

Tension can be calculated by using the equation T = F * L, where T is the tension, F is the force applied to the string, and L is the length of the string.

What factors affect the tension in a string?

The tension in a string is affected by the force applied to it, the length and thickness of the string, and the material it is made of. Other factors such as temperature and elasticity can also impact tension.

What happens if the tension in a string is too high?

If the tension in a string is too high, it can cause the string to break or stretch beyond its elastic limit. This can result in the string losing its ability to return to its original shape and length.

How is tension in a string used in real-world applications?

Tension in a string is used in a variety of real-world applications, such as in musical instruments, suspension bridges, and support cables for buildings. It is also an important concept in the study of mechanics and physics.

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