# Atwood Machine

1. Jan 23, 2005

### extreme

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?

2. Jan 24, 2005

### Staff: Mentor

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 $F_n = (m_2 - m_1)g$.

3. Jan 24, 2005

### extreme

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

4. Jan 24, 2005

### Staff: Mentor

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.

5. Jan 25, 2005

### extreme

equilibrium

Thanks for the quick reply.