Block on a table

  • #1
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[SOLVED] Block on a table

Homework Statement



http://img91.imageshack.us/img91/4285/fudgepq5.th.png [Broken]

I am having some trouble finding the acceleration of the blocks, can anyone help me?

The weight of B is 6 lbs.
The weight of A is 5 lbs.
The friction coef is .3

Homework Equations





The Attempt at a Solution



1) The positive y axis goes from left to right
sum of forces in y = (6/32)*ab = 2T - 1.8

2) positive y axis goes from top to bottom
sum of forces in y = (5/32)*aa = 5-T

I know that aa = -2ab
The way my book derives the relationship between aa and ab is from 'dependent motion' they say that length = sa + 2sb and taking the time derivative twice is 0 = aa + 2ab, the problem is with the way I have my equations setup, when i use aa = -2ab i get the wrong answer, does anyone know how I can fix it so I am using aa = -2ab and getting the correct answer?
I don't see anything wrong with my equations, but i am getting the wrong accelerations. Can anyone help?
 
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Answers and Replies

  • #2
Doc Al
Mentor
44,994
1,269
sign problems

I know that aa = -2ab
Not exactly. You've defined both aa and ab to be positive when B moves right and A moves down, so it should be: aa = 2ab.
The way my book derives the relationship between aa and ab is from 'dependent motion' they say that length = sa + 2sb and taking the time derivative twice is 0 = aa + 2ab,
That's fine, but realize that you are defining the position of B to increase to the right, but sb actually decreases when mass B moves to the right. That's why you need to change the sign.
 

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