Sir Isaac Newton Contest questions

[kingwinner]

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!

 

[Poincare's Stepchild]

Block problem...

You are correct. Total mass = 3 kg. 3 n force gives you 1 m/s*s acceleration. Since this is also the acceleration on m2, with a mass of 1 kg, the force on m2 is 1 n.

Chicken problem...

Concentrate on point C. There are 3 forces, the chicken's weight, straight down, and the two forces (unknown) along the clothes line. Break these two forces into horizontal and vertical components.

Since point C is static...

All the vertical forces sum to 0.
All the horizontal forces sum to 0.

Solve the linear system for the two unknowns.

Handling the pulleys should be easy. Good luck.

BTW, the minimum breaking strength means the line has to withstand the largest tension put on it at any point.

 

[Pete]

There is only 1 string in this system, so the tension force must be the same throughout the length of the string.

If the chicken is gripping the string (which we'll assume she is, otherwise she'd fall off!), then there are effectively two strings, and two tension forces.

 

[kingwinner]

1) Is there a reason as to why the applied force on m1 (3.0N) may not be equal to the force of m1 on m2? At first glance, I thought that m1 is just acting as a mass transferring the 3.0N applied force directly to the next box? So although I got the answer right, I don't get the idea...

2)

If the chicken is gripping the string (which we'll assume she is, otherwise she'd fall off!), then there are effectively two strings, and two tension forces.

Why?

I was told (by my textbook and my teacher) that if there is only 1 string, no matter how you play with it (e.g. putting over a pulley), as soon as it's tight, the magnitude of the tension forces throughout the whole string are equal, even if the direction of the string changes. In this case, there is, indeed, only 1 single string, shouldn't the tension forces pointing away from the chicken be equal in magnitude? (but then the horizontal forces can't balance out)

And for point C, there are 3 forces (2 tension forces, and 1 force of gravity), is this actually the free body diagram for the chicken?

BTW, the minimum breaking strength means the line has to withstand the largest tension put on it at any point.

So "minimum breaking strength" simply equals to the tension force of the string that can be calculated, right?

 

[przyk]

1) Is there a reason as to why the applied force on m1 (3.0N) may not be equal to the force of m1 on m2? At first glance, I thought that m1 is just acting as a mass transferring the 3.0N applied force directly to the next box? So although I got the answer right, I don't get the idea...

If the force m1 applied on m2 were 3.0N, what reaction force would m2 apply on m1? What would the net force and acceleratoin of m1 be?

In this problem, everything is fixed by the fact that m1 and m2 must have the same acceleration.

I was told (by my textbook and my teacher) that if there is only 1 string, no matter how you play with it (e.g. putting over a pulley), as soon as it's tight, the magnitude of the tension forces throughout the whole string are equal, even if the direction of the string changes. In this case, there is, indeed, only 1 single string, shouldn't the tension forces pointing away from the chicken be equal in magnitude? (but then the horizontal forces can't balance out)

You answered your own question here.

And for point C, there are 3 forces (2 tension forces, and 1 force of gravity), is this actually the free body diagram for the chicken?

It'll be the point on the rope the chicken is gripping.

 

[Pete]

I was told (by my textbook and my teacher) that if there is only 1 string, no matter how you play with it (e.g. putting over a pulley), as soon as it's tight, the magnitude of the tension forces throughout the whole string are equal, even if the direction of the string changes.

If you're climbing a rope, is the tension in the rope below you equal to the tension in the rope above you?

 

[kingwinner]

If the force m1 applied on m2 were 3.0N, what reaction force would m2 apply on m1? What would the net force and acceleratoin of m1 be?

In this problem, everything is fixed by the fact that m1 and m2 must have the same acceleration.

You answered your own question here.

It'll be the point on the rope the chicken is gripping.

1) The net force of m1 would be 0, and the net force of m2 would be 3.0N, then! Is that ever possible, physically, for m2 to move and accelerate while m1 is maintaining its state of rest? (i.e. if you apply a force on the first object, the first object stays at rest while the second object, which is in contact with the first object, starts moving and accelerates? Is it possible?)

2) "You answered your own question here."
That's where I think the contradiction comes and I start to get confused...

"It'll be the point on the rope the chicken is gripping."
If I were to draw the free body diagram for the chicken, would there also be 3 forces? (2 tensions and its weight?)

 

[kingwinner]

If you're climbing a rope, is the tension in the rope below you equal to the tension in the rope above you?

Well, I am not sure about that! :m:
But I think the rope below you is loose (not tight) and the rope above you is tight. So if there's a difference in tension, it makes sense.
However, here, the chicken is on the same rope and the whole string is tight...

 

[przyk]

1) The net force of m1 would be 0, and the net force of m2 would be 3.0N, then! Is that ever possible, physically, for m2 to move and accelerate while m1 is maintaining its state of rest? (i.e. if you apply a force on the first object, the first object stays at rest while the second object, which is in contact with the first object, starts moving and accelerates? Is it possible?)

I've never seen it happen. Have you?

2) "You answered your own question here."
That's where I think the contradiction comes and I start to get confused...

You found that the assumption that the tension is the same everywhere in the rope leads to a contradiction. So you reject the original assumption. This technique is used in logic and mathematics a lot (I first saw it formally applied as a proof that sqrt(2) is irrational).

"It'll be the point on the rope the chicken is gripping."
If I were to draw the free body diagram for the chicken, would there also be 3 forces? (2 tensions and its weight?)

The chicken's weight, the upward force the rope is exerting on the chicken, and possibly friction. I'd have to think more carefully about this.

 

[Pete]

If I were to draw the free body diagram for the chicken, would there also be 3 forces? (2 tensions and its weight?)

That's correct. At least, that's the easiest way to tihnk of it.
It might help to pretend the chick is a brick, and the ropes are tied to the brick.

 

[Poincare's Stepchild]

The chicken's weight, the upward force the rope is exerting on the chicken, and possibly friction. I'd have to think more carefully about this.

There wasn't any mention of friction in the problem, i.e. no coefficients of friction. So ignore friction. And everything is static anyway.

BTW, kingwinner, if you try to make the two tensions on the line equal, you can't get the forces to balance at point C.

What you have is that the HORIZONTAL components are equal. The sum of the two VERTICAL components equals the weight of the chicken.

 

[Poincare's Stepchild]

1) Is there a reason as to why the applied force on m1 (3.0N) may not be equal to the force of m1 on m2? At first glance, I thought that m1 is just acting as a mass transferring the 3.0N applied force directly to the next box? So although I got the answer right, I don't get the idea...

The total mass is 3 kg, pushed by 3n. F=ma gives you an acceleration of 1 m/ss.

m2 masses 1 kg, accelerating at 1 m/ss, means a force of 1 n is pushing on it.

Now consider m1. 3 n are pushing from the left. 1 n (the reaction of pushing on m2) is pushing from the right. So a total of 2 n is pushing to the left on m1. Since it masses 2 kg, it is accelerating at 1 m/ss.

Everything works out.

 

[przyk]

There wasn't any mention of friction in the problem, i.e. no coefficients of friction. So ignore friction. And everything is static anyway.

The friction is precisely what prevents the chicken from acting like a pulley and sliding to the middle of the rope, and what makes the the tension different on both sides. Although you don't need to consider it in the problem, it's there.

 

[Poincare's Stepchild]

The friction is precisely what prevents the chicken from acting like a pulley and sliding to the middle of the rope, and what makes the the tension different on both sides. Although you don't need to consider it in the problem, it's there.

I see your point, and agree.