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## Homework Statement

Two 20 kg masses rest on a horizontal table. The bottom mass is connected to a string of negligible mass, which passes over a frictionless pulley with negligible mass and supports a third 20 kg mass that hangs over the side of the table. The coefficient of kinetic friction between the table and the stacked masses is 0.15.

What is the tension in the string?

## Homework Equations

F

_{NET}= ma

F

_{f}= μmg

## The Attempt at a Solution

For the hanging mass:

F

_{w}- F

_{t}= 20a

20*9.8 - F

_{t}= 20a

For the mass on the table:

F

_{t}- F

_{f}= 40a

F

_{t}- 0.15*40*9.8 = 40a

After adding 20*9.8 - F

_{t}= 20a and F

_{t}- 0.15*40*9.8 = 40a

I got 20*9.8 - 0.15*40*9.8 = 60a

so a = (20*9.8 - 0.15*40*9.8)/60 = 2.3 m/s

^{2}

Then I plugged a into 20*9.8 - F

_{t}= 20a to solve for F

_{t}.

F

_{t}= 20*9.8 - 20 (2.3) = 150 N

The answer my book gave me is 137N. Are they asking for something else other than the force of tension of the string?