Pulley with hanging mass

  • #1

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



FNET = ma
Ff = μmg

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?
 

Answers and Replies

  • #2
ehild
Homework Helper
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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:
  • #3
92
0
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:
  • #4
So it's just 9.8*20 - 0.15*9.8*40 = 137 N
Then what did I find?
 
  • #6
ehild
Homework Helper
15,543
1,912
So it's just 9.8*20 - 0.15*9.8*40 = 137 N
Then what did I find?

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
Thank you! I guess the book is wrong.
 

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