# Math Help - Mechanics - elestic string

1. ## Mechanics - elestic string

One end of a light elastic string, of natural length 1.25m and modulus of elasticity 24N, is attached to a fixed point O on a horizontal table. The other end of the string is fixed to a particle of mass 1.6 kg on the table. The particle is pulled along the table until the length of the string is 1.5m.

a) calculate the tension in the string

b) the particle is now released.

i) Given that the table is smooth, calculate the initial acceleration of the particle.

ii) Given that the table is rough and that the particle is on the point of slipping, calculate the coefficient of friction.

OK i think i have only ever answered questions that have inelastic string which is why im struggling.

with an inelastic string the tension would be (1.6) - T = (1.6)(9.8) right?

part bi i think maybe i use the same equation but instead of (9.8) i put 'a' then solve for a?

part bii i think that the point of slipping f = mew*r ??

2. Originally Posted by djmccabie
One end of a light elastic string, of natural length 1.25m and modulus of elasticity 24N, is attached to a fixed point O on a horizontal table. The other end of the string is fixed to a particle of mass 1.6 kg on the table. The particle is pulled along the table until the length of the string is 1.5m.

a) calculate the tension in the string

b) the particle is now released.

i) Given that the table is smooth, calculate the initial acceleration of the particle.

ii) Given that the table is rough and that the particle is on the point of slipping, calculate the coefficient of friction.

OK i think i have only ever answered questions that have inelastic string which is why im struggling.

with an inelastic string the tension would be (1.6) - T = (1.6)(9.8) right?
Where did you get "1.6"? I see no 1.6 in the problem. Where did you get "9.8"? I might guess that is g, the acceleration due to gravity, but since the object is being pulled along a table, there is no gravity in this problem.

The force due to a elastic string is the modulus of elasticity times the amount of stretch: (24)(1.5- 1.25= (24)(.25)= 6N. (Did you copy this correctly? The "modulus of elasticity" is not a force and cannot have units of "N". It should be "Newtons per meter" or "N/m".)

part bi i think maybe i use the same equation but instead of (9.8) i put 'a' then solve for a?

part bii i think that the point of slipping f = mew*r ??
What is "mew"? F=ma. "At the point of slipping", the total force is 0 so the friction force is, in magnitude, the same as the elastic force.

3. Where did you get "1.6"? I see no 1.6 in the problem. Where did you get "9.8"? I might guess that is g, the acceleration due to gravity, but since the object is being pulled along a table, there is no gravity in this problem.
i get 1.6 from the mass of the object. & I thought the string was dangling silly me.

What is "mew"?
mew is coefficient of friction.
F = coefficient of friction x reaction

I did copy it correctly it says 24N :/

So part Bi = Bii ??

4. Hello all
Originally Posted by HallsofIvy
The "modulus of elasticity" is not a force and cannot have units of "N". It should be "Newtons per meter" or "N/m".)
Correction. The modulus of elasticity is a force - it's the force required to double the length of the string.
Originally Posted by djmccabie
One end of a light elastic string, of natural length 1.25m and modulus of elasticity 24N, is attached to a fixed point O on a horizontal table. The other end of the string is fixed to a particle of mass 1.6 kg on the table. The particle is pulled along the table until the length of the string is 1.5m.

a) calculate the tension in the string
So the force required to double the length of this string is 24N. But instead of doubling the length, the length is increased by 0.25m; i.e. by one-fifth of its original length. So by Hooke's Law, the tension is one-fifth of 24N = 4.8N.

b) the particle is now released.

i) Given that the table is smooth, calculate the initial acceleration of the particle.
Force = mass x acceleration

$4.8 = 1.6a \Rightarrow a = 3$

So the acceleration of the particle is $3 \text{ ms}^{-2}$

ii) Given that the table is rough and that the particle is on the point of slipping, calculate the coefficient of friction.
Resolve vertically:

$N = mg=1.6\times9.8$, where $N$ is the normal contact force exerted by the table on the particle.

So if the particle is on the point of slipping, resolve horizontally:

$\mu N = 4.8$

$\Rightarrow \mu = \frac{4.8}{1.6\times9.8}= 0.306$

6. Originally Posted by djmccabie
i get 1.6 from the mass of the object. & I thought the string was dangling silly me.
Oh, I see now. I completely missed that!

mew is coefficient of friction.
F = coefficient of friction x reaction

I did copy it correctly it says 24N :/

So part Bi = Bii ??
No, Bi asks for "acceleration", Bii asks for "coefficient of friction". Completely different questions.

Hello allCorrection. The modulus of elasticity is a force - it's the force required to double the length of the string.
Thanks, Grandad. I was treating the elastic string like a spring and thinking "spring constant"

8. But instead of doubling the length, the length is increased by 0.25m; i.e. by one-fifth of its original length

By looking at the values it is obvious it increases 1/5

but what if the numbers are more complicated? how would i work out the distance?

9. Hello djmccabie
Originally Posted by djmccabie

By looking at the values it is obvious it increases 1/5

but what if the numbers are more complicated? how would i work out the distance?

If the modulus is $\lambda$, the natural length $l$, the extension $x$ and the tension $T$, then Hooke's Law gives:

$T = \frac{\lambda x}{l}$

So in this case, $\lambda = 24, x = 0.25, l = 1.25$ gives

$T = \frac{24\times0.25}{1.25}= 4.8$