Recoiling board
Sam is standing at one end of a 1 0 m board on an ice floor. The board’s mass is 3 0 k g and Sam weighs 5 0 k g .
What distance, with respect to the ice on the ground, does Sam travel if he walks to the other end of the board?
Assumptions : for simplicity, assume that there is no friction between the ice and the board.
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14 solutions
No! I guarantee you that if I walk slowly and carefully on a 3 kg board for 10 meters, let alone a 30 kg one, I could keep it from moving at all and cover all 10 meters, or the vast majority of it. It's all about how slowly and carefully you walk. Were you under the impression that ice is frictionless? "Smooth" per the problem /=/ frictionless. I've been playing hockey on ice for 19 years - I think I know its basic properties.
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You have to realize we are working here with idealized models. All of physics is about simplified models. "Smooth" means, by definition, frictionless in these models. Actual ice (as you know hockey-players know well) is not smooth in that sense; it is only a "rough" approximation of smoothness.
I'd be happy to do this problem again for you with μ k ≈ 0 . 0 2 for wood on ice, if you give me Sam's speed. Or, if you tell me his stride length etc. and both μ k and μ s , I'll give you a more realistic calculation. It's also gonna be a heck more complicated.
This problem was designed to give students simple practice with basic concepts. That ain't "boolsheeit", as you so eloquently state, but good pedagogy. If they try it on real ice, they'll find a model is only a model and therefore limited.
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The word "smooth" is now edited to "frictionless" for further clarity of the model.
I wouldn't bother @Danny Landrum . He's making this kind of remarks on more problems. Dunno what his problem is, but maybe he should invest his time in other things.
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@Peter van der Linden – But is he wrong? Yes, it is simplified, however, we are all students of life. Perhaps he was just making a simple observation about an actual application. If one tried, they could move the board forward by starting slowly up and forward then shifting their weight forward into the decent to stop with added forward momentum on the board, couldn't they?
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@Darwin Simmons – The basic principle is that any change in velocity of the center of mass requires an external force on the system. This must be maintained even in more realistic models.
Here, the system is Sam + plank; and the only possible horizontal external force of any significance is friction exerted by the ice on the plank. If Sam starts walking slowly (acceleration a S ), the plank would "attempt" to slide backward (acceleration a P = a S m S / m P ), but static friction by the ice might prevent this from happening (if a P < μ s ( m S + m P ) g ).
It is also possible to make the board slide forward indefinitely: Sam walks slowly to the end of the plank, relying on friction from the ice to push forward on the plank and keep it in place; then he walks backward quickly , allowing the plank to slide forward; repeat.
Yes I would of thought so to, but we don't have snow is my country so I wasn't sure, and it would be hard for me to test it.
So does this mean the person on the board ended up moving backwards? (in respect to the ice)
Could pls explain how the ratio will differ if the plank had more mass than Sam???? And I had another doubt that if for every 3 meter Sam walks forward, the plank moves 5 meters backward,then isn't he moving closer to the end so shouldn't the denominator be 5-3 ????
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When Sam moves 3 meters forward and the plank slides 5 meters backward, then Sam gets 8 meters closer to the end of the plank.
More formally, m S v S + m P v P = 0 ; v P = − m P m S v S ; v rel = v S − v P = v S + m P m S v S = ( 1 + m P m S ) v S = m P m S + m P v S .
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What do the subscripts s and p represent???Thanks a lot!!!
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@Erica Phillips – S for Sam and P for plank.
If Sam moves 3m forward on the plank, and the plank slides 5m back on the ice, how does that put him an additional 5m closer to the end of the plank? That's not logical. Wouldn't Sam's position relative to the ice be -2m from where he started?
the correct answer would be 0 if the floor was truly frictionless
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Could you explain that?
Not really, here the floor is the ice, and what is frictionless is the two surfaces' system ice-plank.
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Yes, Troy is correct. There is a nonzero coefficient of friction between Sam's shoe soles and the plank - so Sam's shoe soles will exert a horizontal force on the plank in the direction opposite to that which Sam is 'walking' (facing) and the board will move in that direction, uninhibited, because the board-to-ice interface is [perfectly] frictionless. Sam will remain motionless relative to the ice. And pertinent to another reply, conservation of momentum has nothing to do with this.
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@Jesse Otis – When Sam exerts force on the plank, he will accelerate relative to the ice , not just relative to the plank.
Analysis: If Sam exerts force − F on the plank, the plank exerts force F on Sam. They will accelerate (relative to the ice, which is the inertial frame here) at rates a S = − m S F a P = + m P F . Note that a S = 0 if and only if a P = 0 ; the plank can only accelerate if Sam accelerates as well.
Conservation of momentum applies to any physical system on which no net external forces are acting. That is the case here for the Plank + Sam.
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@Arjen Vreugdenhil – I will have to think further about this and get back to you on it. My initial argument is that the plank does NOT exert any force on Sam because the plank is not anchored at all. Thanks for your reply. I will get back to you.
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@Jesse Otis – Think about how Newton's Third Law may apply here.
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@Arjen Vreugdenhil – You are right about a force being exerted on Sam by the plank since he (his shoes) exerts a force on the plank (Newton's third law). But there is absolutely no friction between the plank and the ice. Have you ever roller skated ? Imagine truly frictionless wheels on the skates. You are standing straight and then move one foot backward; you're not going anywhere relative to the ground (unless it is downward when you fall from having the moved foot out from under you). That is exactly what this situation is; the roller skate shoes are fastened to the person's feet (can liken that to a large coefficient of friction); foot exerts a rearward force on skate, skate exerts a forward force on foot. Skate wheels roll back with no friction at all between them and the floor. Skater doesn't go anywhere with that method; skater has to orient skates so that the wheels 'push' against the floor in a sideways fashion.
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@Jesse Otis – Rolling motion is not exactly frictionless; in fact, it only works because of friction. If there were no static friction between floor and wheel, the wheel would not roll but slip.
But how about this: Let's put Sam on roller skates. As you describe in your last sentence, in order to skate on the plank, Sam has to push sideways (and partially backwards) on the plank. This makes Sam go forward and the plank go backward.
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@Arjen Vreugdenhil – Are you currently teaching ? If so, what subjects and at what level (e.g. university) ?
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@Jesse Otis – I used to teach Physics at high school / undergraduate college, and Chemistry at high school. Currently I am teaching Theology at a college.
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@Arjen Vreugdenhil – Awesome; no wonder you know so much about physics & mathematics. I bet that you could create and teach a course called Mathematical Theology. :-) Thanks for your patience with me on that Sam the plank walker problem.
@Arjen Vreugdenhil – Arjen -- Please let me know if you think my analogy of roller skating is accurate with regard to the guy standing on the plank which is on ice. I would like to know if I am off-base or correct in my thinking on that. How can we find out for sure ?
Thanks I couldn't figure it out, and your explanation helped me understand
Hello everyone! I am a student who is still learning. Can somebody please reply to me on how to get it in the simplest possible solution, as soon as possible?
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i did this by reenacting the scenario with my hand and ballpen on the table. silly but it worked xD
sorry sir....!!!! but would you like to explain how it done . actually i can't understand ....i am student don't know this type of thinks thanks you sir your faithful MD ajam
funny but i solved this by reenacting/visualizing it by assuming that my 2 fingers=sam walking and my ballpen=10m board and the table as the ice on the ground. it worked lol
Since there is no external force, Centre of mass (CoM) remains at same place as Sam reaches the other end and the board slides back. Initially the CoM was at 30*5/(50+30) = 15/8 m from the left end of the board. In the end position, the CoM would be at 15/8 m from right end by symmetry. Hence the total distance moved by Sam relative to ice is 15/8+15/8 = 30/8 = 3.75m
assuming the fact that we can put real variables such as friction into the equation and how much the board weighs, he can walk more than 5 meters without the board actually moving.
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when the problem says smooth, it is implied that the coefficient of friction of the ice is 0. I made the same mistake taking friction into account...
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The word "Smooth" is now replaced by "Frictionless" for further clarity.
"since there is no external force" you mean "since the external forces balance each other..." (ie weight + reaction of ice, both vertical, make vector zero) "... the situation is the same as if there was no external force".
Indeed, even if external forces balance each other (nth law of Newton... 1st?), they exist (effects : deformation of objects, compression, elongation... ; even also changes of length if we would consider relativist physics...). We do not take them in account in this exercise.
The center of mass of Sam and the board is 3 0 + 5 0 3 0 from where Sam is to the center of the board, or about 1 . 8 7 5 meters. This center of mass will not move during Sam's walk, but he ends up on the other end of the board. Hence, he will have travelled relative to the ground twice that, or 3 . 7 5 meters, or less than 5 meters.
Did the same
Why 30/(30+50)? Is it actually (50 * 0 + 30 * 1)/(30+50)? I don't understand
I thought about it intuitively. We can think of mass as how hard it is to move an object. Since Sam is heavier than the board he is harder to move. If they had the same mass Sam would travel the same distance as the board (10 meters are to be travelled so Sam would go 5 meters). Since Sam is heavier he would travel less than 5 meters.
I agree with you
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Thanks!
If Sam was weightless, he would walk the full length of the board, 10m. If he weighed the same amount as the board, he would make it halfway, 5m. Since he weighs more than the board, he doesn't make it to halfway.
Assuming a sufficiently low friction between the ice and the floor, as Sam begins to move forwards of the board, an equal and opposite force will slide the board backwards. At the end of Sam's walk, the combined centre of mass of Sam and the board, which started slightly more than half way towards Sam’s centee of mass (less than 2 . 5 m then the combined centre of mass will be in the same place, due to inertia, but Sam will be at the opposite end of the board, and so will have moved slightly less than 5 m from their original position
@Mike Pannekoek : Precisely for the reason that center of mass(COM) stays put, Sam has to walk 8 2 5 from left to 8 2 5 right of COM. Accordingly, he walks slightly more than 5m, 6 . 2 5 m to be precise.
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shouldn't the centres of mass be less than 2.5m apart?
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(the combined centre of mass and Sam's)
Yes you are right.
If sam move carefully enough wouldn't he be able to move from one side of the plank to the other without breaking the coefficient of static friction between the plank and the ice?
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The question does not state that the ice is frictionless, so you are more or less correct.
Yes, that's correct, Dean. "Smooth" /=/ frictionless.
"Assuming a sufficiently low friction between the ice and the floor" - Why would you think that you can assume anything? The problem said the ice was "smooth" not frictionless. If it was frictionless, they would have said "frictionless", not specifically chosen to use the word "smooth" instead of frictionless. I guarantee you that if I walk slowly and carefully on a 3 kg board for 10 meters, let alone a 30 kg one, I could keep it from moving much at all and cover all 10 meters, or the vast majority of it. It's all about how slowly and carefully you walk. I've been playing hockey on ice for 19 years - I think I know its basic properties.
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keep in mind that the problems of the week are community made, and only have as much quality control as the person who originally posted it.
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Smooth ice is very slippery and has a very small coefficient of friction. Even if it is not frictionless but very smooth, the answer will still be the same, but the exact calculations will be very complex.
For further clarity, the word "Smooth" is replaced by "Frictionless".
Are we seeing different problems? Mine definitely says "frictionless".
The given assumption falls short. Its purpose is to isolate the problem from all unknowns relevant to the problem, but these not only comprehends friction between ice and board, but also other friction forces like air resistance, and other minor interactions with the environment, like... breath. We must assume that Sam breath is done without incurring into net forces applied to the Sam+Board system, considering that we want to assume there is no friction involved (we might discuss if it is enough to demand the same air velocity in Sam's inhale and exhale).
With all these assumptions, the system Sam+Board is an isolated one, with its mass center always in rest. If Sam moves, the board must also move, in such a way that the mass center stays fixed. So, in order to answer the question, we only need to calculate the mass center in two situations:
When Sam is at the left: 8 0 k g 5 0 k g × ( − 5 m , A ) + 3 0 k g × ( 0 m , B ) = ( − 5 m × 8 5 , 8 5 A + 3 B )
When Sam is at the right: 8 0 k g 5 0 k g × ( + 5 m , A ) + 3 0 k g × ( 0 m , B ) = ( + 5 m × 8 5 , 8 5 A + 3 B )
where we defined the origin at the center of the board. This mass center is fixed in relation to the ice. We are only interested in its horizontal coordinate. In the beginning, Sam is 5 − 2 5 / 8 = 1 . 8 7 5 m to the right of that center. After, it is the same value to left. So, he moves in relation to the ice, 3 . 7 5 m .
If he walks the board, the board will slip back because he weighs more than the board. If he weighed the same as the board he would walk only 5m. So since he weighs more than the board he will walk less than 5m.
If there is no friction between the Ice-Pad and the board he, relatively to the pad travels 0 meters. Only the board moves. Isn‘t this right ?
I agree with you. The masses of Sam and the board are irrelevant because their (gravitational) force is downward. As Sam walks (the plank :-) ), the force of his steps is horizontal pushing the board to the left while he remains stationary (travels 0 meters) with respect to the ice floor. I think too many folks are over thinking this....
With no other forces involved, the center of mass (COM) between Sam and the board will not change it's location relative to the ice. That COM is somewhere between Sam's COM and the board's COM, but closer to Sam because he weighs more than the board.
When Sam reaches the end of the board, the COM of both hasn't changed but he is on the other side of it. With the board's COM being 5m from Sam before and after, he has effectively walked less than half of 5m twice over relative to the ice. Therefore, he has walked <5m relative to the ice.
We don't need a specific answer. We just need to know if it is exactly 5, or less, or more. 5 is half the length. So, we observe. If the mass (Sam vs board) is the same, it will be exactly 5. Because Sam weight more, the board pushes to Sam's back more, so Sam will travel less, giving the answer is less than 5
Conservation of linear momentum demands that the initial momentum be equal to the final momentum in a "frictionless" system. In our system we have 0 initial momentum, meaning, our system should behave in a way as to always keep the total momentum, mv, 0. Assuming Sam has a velocity of 1m/s and takes 1 step in a second (note that the velocity only dictates the duration of the process of reaching the other end and not the relative distance travelled, hence 1m/s is reasonable for ease of analysis), for each step Sam takes, the rod moves 5/3 (1.66) meters in the opposite direction underneath his feet, which means he moves 2.66 meters across the rod with each step. 10/2.66 = 3.75. Sam moves only 3.75m relative to the ice.
La madre estaba deslizante, entonces cuando sam se movio en vez de crear dezplasamiento la tabla fue la que fue arrastrada hacia atras. :V
Since the ice is perfectly smooth, the board(and San along with it) will move backwards continuously. Therefore, in respect to the ground, he moved backwards.
I'm not sure if the ice mattered, because I don't think that's the correct way to get the answer.
Conservation of momentum means that the plank and Sam move relative to the ground at a speed ratio of 3:5. Thus for every 3 meter Sam walks forward, the plank moves 5 meters backward. We see that distance Sam gets closer to end of plank distance Sam walks relative to ground = 3 + 5 3 = 8 3 < 2 1 . Therefore Sam walks less than half of the length of the plank, i.e. less than 5 meter.
Note that if Sam and the plank had equal mass, the answer would be half of the length of the plank (5 meter); and if the plank had more mass than Sam, the answer would be greater than half of the length of the plank.