# Find Acceleration of Blocks on Table [SOLVED]

• jesuslovesu
In summary, the conversation is about finding the acceleration of blocks A and B on a table, with weights of 5 lbs and 6 lbs respectively and a friction coefficient of 0.3. The equations for sum of forces in the y-axis are discussed, with the problem being the relationship between aa and ab. The issue is resolved by changing aa = 2ab to aa = -2ab to account for the direction of motion.
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

## 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?

Last edited by a moderator:
sign problems

jesuslovesu said:
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.

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!

## 1. How do you find the acceleration of blocks on a table?

The acceleration of blocks on a table can be found by using the equation a = F/m, where a is the acceleration, F is the net force acting on the blocks, and m is the mass of the blocks. This equation is based on Newton's Second Law of Motion.

## 2. What is the difference between static and kinetic friction?

Static friction is the force that prevents an object from moving when a force is applied to it, while kinetic friction is the force that acts on an object that is already in motion. The magnitude of static friction is always equal to or greater than the magnitude of kinetic friction.

## 3. How do you calculate the net force on the blocks?

The net force on the blocks can be calculated by finding the vector sum of all the individual forces acting on the blocks. This can be done by using the formula Fnet = ΣF, where Fnet is the net force and ΣF is the sum of all the forces.

## 4. What factors affect the acceleration of blocks on a table?

The acceleration of blocks on a table can be affected by the net force acting on the blocks, the mass of the blocks, and the coefficient of friction between the blocks and the table. Other factors such as air resistance and the shape of the blocks can also have an impact on the acceleration.

## 5. Can the acceleration of blocks on a table be negative?

Yes, the acceleration of blocks on a table can be negative. This would indicate that the blocks are slowing down or moving in the opposite direction of the applied force. It is important to consider the direction of the net force and the initial velocity of the blocks when determining the sign of the acceleration.

• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
7K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
3K