Atwood Machine simulation problem

AI Thread Summary
In the Atwood Machine simulation, m1 is 1.0 kg and m2 is 1.1 kg, with m2 resting on the floor. The tension in the rope at 2.54 seconds is 9.789 N. The equilibrium condition is established with the equation Fn + T - m2*g = 0, leading to a calculated normal force (Fn) of approximately 1.0 N. The discussion confirms that the provided calculations are correct under the assumption of equilibrium. The normal force and tension values are crucial for understanding the system's dynamics.
extreme
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Need a check on this one please. In an Atwood Machine simulation, m1 is 1.0kg and m2 is 1.1kg. m2 rests on the floor that exerts a normal force, Fn, on m2. when the system is in equalibrium, the tension, T, in the rope at 2.54s is 9.789. If Fn +T- m2*g = 0 then Fn = .991. Is this correct?
 
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equilibrium

If the question is "What normal force does the floor exert on m2 when m2 rests on the floor?" then you're probably OK. (Although I have no idea what 2.54s has to do with anything.)

The forces on m2 are: F_n + T - m_2 g = 0; similarly, the forces on m1 are: T - m_1 g = 0. So F_n = (m_2 - m_1)g.
 
equilibrium

The question "What is the normal force, FN, and what is the tension, T, in the rope? T (at t = 2.54s) = ______. At equilibrium: FN + T – m2 • g = 0. Then FN = _______." This is a timed simulation and at 2.54s the rope tension,T, is 9.789
 
extreme said:
The question "What is the normal force, FN, and what is the tension, T, in the rope? T (at t = 2.54s) = ______. At equilibrium: FN + T – m2 • g = 0. Then FN = _______." This is a timed simulation and at 2.54s the rope tension,T, is 9.789
OK, now I see what you're talking about. Since T is given and equilibrium is assumed, your answer is correct; rounded off, the normal force would be 1.0 N.
 
equilibrium

Thanks for the quick reply.
 
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