Find Acceleration of Blocks on Table [SOLVED]

[jesuslovesu]

[SOLVED] Block on a table

Homework Statement

http://img91.imageshack.us/img91/4285/fudgepq5.th.png

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?

 

[Doc Al]

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.

 

[Moetasim]

First, it is important to note that the question has been marked as [SOLVED], so it is likely that the original poster has already found the answer to their problem. However, for the sake of providing a response, I will give some suggestions for finding the acceleration of the blocks.

To start, it is important to identify the forces acting on each block. For block A, we have its weight (5lbs) acting downwards and the tension force (T) acting upwards. For block B, we have its weight (6lbs) acting downwards, the friction force (0.3*6lbs) acting in the opposite direction of motion, and the tension force (T) acting upwards.

Next, we can use Newton's second law (F=ma) to set up equations for each block. For block A, we have 5-T=5a (a is the acceleration of block A). For block B, we have 6-0.3*6-T=6a (a is the acceleration of block B). We also know that the blocks are connected by a string, so they must have the same acceleration (a).

Now, we can solve for the tension force (T) by setting the two equations equal to each other: 5-T=5a and 6-0.3*6-T=6a. This gives us T=5-5a and T=6-6a. Setting these equal to each other, we get 5-5a=6-6a, which gives us a=1. This means that the blocks are accelerating at a rate of 1 m/s^2.

I am not sure what equations you were using in your attempt at a solution, but it is possible that there was a mistake in your setup or calculations. It is always important to double check your work and make sure that all equations and values are correct. I hope this helps!