0
$\begingroup$

In space for example if we apply two forces 10,6 respectively at the end of an elastic string (10 at the right end and 6 at the other) then obviously that body will expand and accelerate. But how would you find young's modulus $ {Y}= \frac {FL}{Al}$ ? What would be that restoring force in the string ? Both acceleration and expansion in a body makes the problem really difficult for me . I thought several ways to solve but couldn't. Though I heard that this situation is similar to a string suspended from the ceiling and someone applying force at the free end and also that the restoring force is the average of both force e.g. (10+6)/2 in this case. I don't know wether it's correct. However I'm looking for an intuition to solve this problem and to write an expression for young's modulus.

$\endgroup$

2 Answers 2

0
$\begingroup$

The 2 forces will result in a nett force of 4 in the direction of the '10' force.

This will cause an acceleration given by a = F/m based on the system mass.

As for the elastic - it will be subject to a stretching force of 4 and will expand according to the spring modulus.

Once the extension has reached stability then the system will accelerate at the 'stretched' length 'indefinitely'

$\endgroup$
0
$\begingroup$

Suppose a mass $m$ hanging by a rope inside a rocket at rest on earth, attached to the top. The weight $mg$ is balanced by the top upward force $F_u$.

If the rocket is launched and have now an acceleration $a$, the Newton's second law states that: $F_u - mg = ma$.

In order to know the elastic effect we note that in the first case the tension in the rope is $F_u = mg$, but in the second case it is $F_u = mg + ma$. A way to understand it is to imagine staying on the accelerated frame. There, the total downward force is $mg + ma$, there is a fictitious $ma$ force added to the weight.

In the OP example, the Young module is $$Y = \frac{10}{\Delta}$$ where $\Delta$ is the elongation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.