Tug of War!!

This is an interesting problem I came across in the book "The Concepts of Physics"by H.C.Verma.

"Can you EVER win a tug of war on a smooth frictionless surface?"

(Note : The winning line is exactly in the centre of the line joining the two teams(assume ,for simplicity, only two people playing against each other). Also, the masses of the opposite players need not be equal.)

#Calculus

Easy Math Editor

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Comments

Rockets? You may try to blow in the opponent's direction; the one with more breath would win.

A Former Brilliant Member - 7 years, 6 months ago

Will you please elaborate, Yash Talekar?

SAMARTH M.O. - 7 years, 6 months ago

There is no friction in space, yet rockets move forward. In theory you could use the same principle...

A Former Brilliant Member - 7 years, 6 months ago

Well.....I think its time I tell my answer.. The answers given in the comments strongly use the floor as the only initial potential source for finally pulling the other person.

But my answer is different....

My answer is that one will always win the match if and only if he/she is more massive than the opponent.

Explanation : Consider the two opponents and the rope as your system. There are no net external forces on the system(as gravity is undone by normal reaction). Consider that you are the more massive opponent. The winning line is exactly midway between you and your opponent. But, the centre of mass of the system is closer towards you(the more massive one) ,that is, the c.o.m is somewhere between you and the winning line.

Now, let the match start. Whatever be the interactions, the force on you and the opponent will be equal(of course considering the rope to be massless and inextensible). So, the centre of mass does not move. So, whatever you guys do, you both have to end up at the centre of mass, which is inside your domain. So, your opponent has to cross the mid-line. AND YOU WIN!!

So, it does not depend who does what. The more massive one would always win..

SAMARTH M.O. - 7 years, 6 months ago

din get u .. !!!

Ramesh Goenka - 7 years, 6 months ago

frictionless surface pe th kahan se jeet paoge .. force hi laga nai payega dunon mein se koi .. !! .. !!! the reason being that ki agar thda sa bhi vertical ke saath angle bana toh slip ho jayenge.. !! but haan bina angle banaye agar force possible hai toh jisne zyaadaa force lagaya vh jeet gaya.. !! but as for the fact the surface is frictionless so answer being no.. !!

Ramesh Goenka - 7 years, 6 months ago

HCV uncle toh genius hai......the answer is a big NO.....COM ka concemt ghusane ki koi zarurat nahi hai...no force no acceleration.....no horizontal component.....

Max B - 7 years, 1 month ago

Max B- well, pal, you seem to have taken the net force on the whole system. It is true that the net force on the whole system is zero. But please understand that zero net force on the system does not imply that there are no internal forces at all! Its true that the COM will not accelerate. (and further, remains at rest due to zero initial velocity) It is EXACTLY due to this fact that players have to finally reach their COM.

Here, it is absolutely true that each player will experience a force. Do not mess up with the action-reaction pairs.

(And please try to write in English. I am sorry, but the spelling of hindi words confuse me. )

SAMARTH M.O. - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$