# Tension in the atwood machine

1. Jul 26, 2009

### toshiba_girl

1. The problem statement, all variables and given/known data

For this problem the atwood's machine is used. There is a simulation where, m1 is 1.0 kg and m2 is 1.1 kg. Mass m2 rests on the floor that exerts a normal force, FN, on m2. There is no motion (i.e.. the system is in equilibrium). What is the normal force, FN, and what is the tension, T, in the rope? T (at t = 2.54 s) = ______. At equilibrium: FN + T – m2 • g = 0. Then FN = _______. For the simulation, at t=2.54 s, the tension of the pulley system is stated to be 9.789 N. Therefore I was wondering whether this value for tension is the value to be used or whether tension in the rope is to be calculated by other means.

2. Relevant equations

FN + T - M2 * g = 0

3. The attempt at a solution

I was thinking that maybe FN = (1.1 * 9.8) - (1.0*9.8) = 0.98 N
and T = (1.1 * 9.8) + (1.0 * 9.8) = 20.58 N

However this response does not seem to be correct to me. Can anyone please help me out. Thanks

2. Jul 26, 2009

### rl.bhat

At t = 2.54 s both mass are moving with some acceleration. Both will be moving with the same acceleration. m2 must be moving in the downward direction.
The acceleration of m2 is given by (m2*g - T)/m2
The acceleration of m1 is given by ( T - m1*g )/m1.
Equate them and find the tension in the rope.

3. Jul 27, 2009

### toshiba_girl

Ok but then how would i calculate the normal force exerted on m2?

4. Jul 27, 2009

### PhanthomJay

You seem to have 2 parts to the problem. The first asks for the normal force of the floor on m2, when the system is in equilibrium. The second apprently gives you a tension force when m1 is lowered by some force?? I don't know about part 2. But as for part 1, the equilibrium case, look first at the forces acting on m1. That will allow you to solve for the rope tension. Then you can use that value to solve for Fn acting on m2.