Sir Isaac Newton Contest questions

kingwinner · 04-04-2006

The following are 2 of the Sir Isaac Newton Contest questions in the past years that I found in my textbook. Unfortunely, I am having a fair amount of problems and questions with these. It would be nice if somebody can explain, I would appreciate!

1)

First, I have assumed that the applied force on m1 is not equal to the force of m1 on m2, I then got the force of m1 on m2 (or the contact force) is 1.0N, is this correct? (there is no answer in the textbook)

Second, I would like to ask why wouldn't the applied force on m1 be equal to the force of m1 on m2? Shouldn't m1 be just acting as a mass transferring the 3.0N applied force directly to the next box? I don't get the idea...

2)

This question I am having a lot of troubles with! Is the string fixed to the poles on both ends? But it seems that there are pulleys around, is the string fixed to the pole below the pulley?

There is only 1 string in this system, so the tension force must be the same throughout the length of the string. So there are 2 tension forces pulling in opposite directions along the string at point C, but since the angle aren't equal, which, with the force of gravity, should balance out to zero. However, the horizontal forces don't balance out, this is where I got stuck...

And finally, what does minimum breaking strength actually mean? I am not sure what the question is looking for!

Pyrotex · 04-06-2006

Quote:

Originally Posted by kingwinner

...First, I have assumed that the applied force on m1 is not equal to the force of m1 on m2, I then got the force of m1 on m2 (or the contact force) is 1.0N, is this correct? (there is no answer in the textbook)

Let's assume the applied force on m1 by m2 (and on m2 by m1, for they ARE equal) is a million N. Obviously wrong, but why? How the hell could you be applying 3.0 N to the first block if its force on the second is so huge? If a million N is obviously too big, what is the answer? 3.0 N. Same as your force on the first block.

Quote:

Originally Posted by kingwinner

......This question I am having a lot of troubles with! Is the string fixed to the poles on both ends? But it seems that there are pulleys around, is the string fixed to the pole below the pulley?...And finally, what does minimum breaking strength actually mean? I am not sure what the question is looking for!

If you tied the string to the chicken and pulled it up, letting the chicken just hang from the string, then the minimum breaking strength (mbs) of the string must be equal to the weight of the chicken -- the mbs/string would be 2.0 N.

No, the tension in both arms of the string may not be the same. Ignore the pulleys -- the string is essentially fixed at each end.

I would solve it by taking the vertical and horizontal components of the forces.
Fv (of chicken's weight) = Facv + Fbcv (sum of the vertical comps in strings)
or Fac (sin 30) + Fbc (sin 45) = 2 kg * G = 2 N
.5 Fac + .71 Fbc = 2

The chicken isn't moving, so Fach = Fbch
(rather than Fac = Fbc !!!!! This is the sticking point!!!)
or Fac (cos 30) = Fbc (sin 45)
.87 Fac = .71 Fbc

We have
Fac + 1.41 Fbc = 4
Fac = 4 - 1.41 Fbc
and
Fac = .71/.87 Fbc = .82 Fbc

substituting
4 - 1.41 Fbc = .82 Fbc
0.6 Fbc = 4
Fbc = 6.7 N
Notice this is way more than the weight of bird.

Fac = .82 Fbc = 5.4 N

The mbs/string must be 6.7 N, or otherwise the short stretch would break.

Now, convince yourself that the tension in the two parts of the string do NOT have to be equal. Do this by exagerating the problem. Try angles of 1 degree and 89 degrees. Remember, the chicken is holding onto the string, and not moving -- ie, in equilibrium..

Qfwfq · 04-07-2006

Quote:

Originally Posted by Pyrotex

If a million N is obviously too big, what is the answer? 3.0 N. Same as your force on the first block.

Now that deserves a good whack on the knuckles!

It says frictionles table so they get accelerated. If you want to help King with it, think more carefully and remember your freshman course exercises.

sigsfried · 04-07-2006

Q1. The total mass is 3kg
The acceleration in 1m/s
The force required on the second one is therefore 1N. Otherwise the second block would accelerate at a different speed to the second one.

Qfwfq · 04-07-2006

Quite right. The two accelerations couldn't be different, in steady force regime.

sigsfried · 04-07-2006

Quote:

Originally Posted by Qfwfq

Quite right. The two accelerations couldn't be different, in steady force regime.

Wouldn't matter is it wasn't a steady force. If the two accelerations are different one of two unphysical situations occur. Bock one passes through block two or block two seperates in which case how is it still accelerating. This is true unless block 1 is slowed.

Qfwfq · 04-07-2006

A knock can propagate through solid objects in contact, with the last one getting projected away. The acceleration lasts briefly. This scenario, called impulsive regime, requires a duration comparable to the time it takes to propagate through the solid object.