Welcome to NRICH.

 
Hooke's Law
By Anonymous on Thursday, June 15, 2000 - 09:06 pm:

Please could someone help me solve the following Hooke's Law problem.

A ball of mass 0.5kg is attached to one end of a cord of unstretched length 0.6m whose other end is fixed. When a horizontal force of magnitude Q N is applied to the ball, holding the ball in equilibrium, the cord increases in length by 10% and is inclined at an angle arcsin 3/5 to the downward vertical. Calculate the value of Q and the modulus of the cord.

I know the change in length (x) is 0.06m but don't seem to be able to find the tension (T) after resolving. Without the tension, I can't do the rest.

Thanks to anybody that can help
By Andrew Smith (P2517) on Thursday, June 15, 2000 - 10:24 pm:

If the tension = T N, then the horizontal component of tension is 3T/5 N and the vertical component is 4T/5 N.
The vertical component = Weight = 0.5g (g=gravity)
The horizontal component = Q
0.5g=4T/5 means T= 5g/8 N
Q=3T/5=3(5g/8)/5=3g/8 N

T=fx/l (f=modulus of elasticity)
5g/8=f×0.06/0.6
f=25g/8 N

Hope that helps

By Anonymous on Friday, June 16, 2000 - 08:46 pm:

Thanks Andrew, that helps a lot. However, I'm having a problem understanding the reasoning with another Hooke's Law question (I'm new to this topic I'm afraid):

Two elastic strings both of natural length 1.5m and moduli 8N and 6N respectively are joined together at both ends. One end of the combined string is fixed and a particle of mass 2kg is attached to the other end and hangs in equilibrium with the string vertical. Calculate the tension in each string.

Now, all previous questions I have answered made use of the fact that in equilibrium, the tension in the string is equal to the weight of the particle (in this case 2g). But this question wants the tension in both strings. I now know that it is the combined tensions in both strings that equal the weight, in the ratio 4:3. However, I'm having trouble understanding why this is the case. I still keep thinking the tension anywhere on both strings is 2g for some reason.

Maybe you could explain this better than the book does.

Thanks
By Andrew Smith (P2517) on Friday, June 16, 2000 - 10:05 pm:

I agree that is a bit of a confusing concept. I can explain it but I don't know if it will be any clearer than the book!

The sum of the tensions in both strings will be equal to 2g by resolving vertically.
Also the extension in both strings is equal, call it x.

So T1=8×x/1.5 , T2=6×x/1.5 and T1+T2=2g.

Therefore you can find the extension and thus the tensions or just say;

T1:T2 = 8×x/1.5:6×x/1.5 = 8:6 = 4:3, avoiding the calculations.

I can give you a different example which may help to convince you, take two strings of equal length 1m, one having modulus 100000000N and one having modulus 1N. Attach a weight of 2kg to the bottom of the string of modulus 100000000N and it will hardly extend and have tension=2g. Now attach the other string to the top and bottom and it has very low tension as it has virtually no extension (it's like it is hanging freely with no weight on it) while the other string retains a tension near to 2g. Sorry if this isn't very clear but I hope it helps to show why they don't have equal tensions.

Andrew

By Katharina JüRges (P2657) on Sunday, July 2, 2000 - 01:30 pm:

Although the question has been posted more than two weeks ago I hope you will read my answer, however.

You said that you still kept "thinking that the tension anywhere on both strings is 2g for some reason". Well, I will show you that this can't be true.

The weight of the mass which is attached to the two strings is a force which acts downwards. The tension in the two strings which is a force as well has to compensate the weight, so that the whole thing can be in equilibrium.

This is why the sum of the tensions in the two strings must be 2g, not each tension.

If the tension in both strings were 2g then the force acting upwards would be the double of the force acting downwards, and so there would be no equilibrium at all, but the mass would be pulled upwards.

I hope this was somewhat clear. Actually, I had Hooke's Law three or four years ago and I never heard of moduli, but this was certainly because I went to school in Germany at that time and we didn't treat the subject that profoundly.

However, what I just explained to you is based on another law of Physics, and you can trust that my reasoning was correct.