# Pulley with hanging mass

1. Dec 3, 2011

### l607309608126

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

2. Relevant equations

FNET = ma
Ff = μmg

3. The attempt at a solution

For the hanging mass:
Fw - Ft = 20a
20*9.8 - Ft = 20a

For the mass on the table:
Ft - Ff = 40a
Ft - 0.15*40*9.8 = 40a

After adding 20*9.8 - Ft = 20a and Ft - 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/s2

Then I plugged a into 20*9.8 - Ft = 20a to solve for Ft.
Ft = 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?

2. Dec 3, 2011

### ehild

Are those blocks on the top of each other and accelerating together? Is the coefficient of friction between the blocks not specified?

ehild

Last edited: Dec 3, 2011
3. Dec 3, 2011

### TaxOnFear

I got what your answer book gives. I think you're over complicating the question.

You have two forces in play that you need to take into consideration:
The weight of the hanging mass, and the frictional force due to the hanging mass.

Last edited: Dec 3, 2011
4. Dec 3, 2011

### l607309608126

So it's just 9.8*20 - 0.15*9.8*40 = 137 N
Then what did I find?

5. Dec 3, 2011

Correct.

6. Dec 3, 2011

### ehild

You have found the resultant force acting on the system of masses. The acceleration is 137/(3*20)=2.28 m/s2. That is the same you got.
The tension in the string is 20g-20a=150 N. The book is wrong.

ehild

7. Dec 3, 2011

### l607309608126

Thank you! I guess the book is wrong.