two people- one twice as heavy as the other one, play tug-of-war with a 12m ideal rope while standing on a frctionless ice surface.Would any of them move and if he does, how far?explain why.
Their center of gravity would not move because there is no net force on the system comprising the two masses.
Say they start a distance d apart
the 2m mass is at x = 0
the 1 m mass is at x = d
Then the center of gravity is at
2m Xcg = m (d-Xcg)
2 Xcg = d - Xcg
Xcg = d/3
which is where it stays.
In this scenario, the two people are standing on a frictionless ice surface, and they are playing tug-of-war with a 12m ideal rope. One person is twice as heavy as the other person. Let's analyze what would happen in this situation.
In a tug-of-war, the outcome is determined by the net force acting on each side. The net force is the difference between the forces applied by each person. In this case, the person who is heavier exerts a greater force.
Since the surface is frictionless, there are no additional forces to consider. The only forces acting on the system are the forces applied by the individuals in the tug-of-war.
Due to Newton's third law of motion, the force exerted by one person on the other person is equal in magnitude but opposite in direction. So while the heavier person exerts a greater force, it is countered by an equal force in the opposite direction from the lighter person.
Now, let's consider the motion. If the net force is zero, according to Newton's first law of motion (the law of inertia), the two people will not move. This happens when the force exerted by the heavier person is balanced by the force exerted by the lighter person.
In this case, the heavier person exerts a greater force, but they are also twice as heavy as the other person. Therefore, the heavier person's force is mitigated by their greater mass. The net force is zero, and both people remain stationary.
In summary, since the net force in the tug-of-war scenario is zero, neither person will move on the frictionless ice surface.