Atwood Machine In this simulation m1 is 1.0 kg and m2 is 1.1 kg. Mass m2 is 1.1

Question

Atwood Machine

In this simulation m1 is 1.0 kg and m2 is 1.1 kg. Mass m2 is 1.1 kg. Mass m2 rests on the floor that exerts a normal force FN, on m2 * g = ). Then FN = ? and T (at t=2.54s) = I pluged everything into the equation..

mCg = (1.0kg)(9.8m/s^2) = 9.8N

mEg = (1.1kg)(9.8m/s^2) = 10.78N

FT - mEg = mEaE = -mEa

FT - mcg = mca = +mca

(mE - mc)g = (mE+mc)a

a = (mE - mc) /( mE + mc)g = 1kg - 1.1kg / 1kg + 1.1kg g

a= -0.048g = - 0.47m/s^2

and then

FT = mEg - mEa = mE(g - a)

= 1.0kg((10.78m/s^2 - (-0.47g)) = 11.25N

= 1.1kg((9.8m/s^2 +.(-47g)) = 10.26

And thats where is doesnt work out any help would be great.

Explanation / Answer

First of all the value of the acceleration would be positive because
ME = 1.1 Kg and Mc = 1 kg
therefore a = 0.47 m/s2
Now equation of mass E
FT = ME (g - a) = 1.1(9.81 - 0.47) = 10.274 N
Equation of mass C
FT = MC(g+a) = 1(9.81 + 0.47) =10.28 N

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