Assuming acceleration is constant

In summary, the conversation discusses two physics problems, 12.126 and 12.72, that involve determining the acceleration of blocks with constant forces applied. The difference between the two problems is that in 12.126, the cables are positioned at a constant angle, allowing for a constant acceleration, while in 12.72, the angle changes and the tension in the cables also changes, resulting in a non-constant acceleration. The conversation also addresses the importance of considering the direction of acceleration and how it affects the solution to the problem.
  • #1
cipotilla
31
0
I have posted 2 problems, 12.126 and 12.72. I have read in the book that you can assume the acceleration remains constant when the forces applied are constant.

In problem 12.126 that is the case, the weights of the blocks and the forces P and T remain constant so I assumed that the acceleration of the blocks were also constant and I solved the problem successfully.

But in problem 12.72, I obviously cannot assume the acceleration is constant, why not? I am neglecting friction and the weights of the blocks are contant, the tension in the cord is also constant, isn't it? What is the difference between these problems?
 

Attachments

  • Assuming constant acceleration.pdf
    34.6 KB · Views: 672
Physics news on Phys.org
  • #2
cipotilla said:
But in problem 12.72, I obviously cannot assume the acceleration is constant, why not? I am neglecting friction and the weights of the blocks are contant, the tension in the cord is also constant, isn't it? What is the difference between these problems?
As block A moves, the angle that the cord makes will change. This will affect the forces involved. (And don't assume the tension in the cord remains the same as the angle of the cord changes.)
 
  • #3
As the angle changes, I thought that the components of the tension force would change but the resusltant T wouldn't. Why is this incorrect?
 
  • #4
Why do you think that?

Did you actually write down the equations and solve for the tension? Does it depend on theta?
 
  • #5
I solved problem 12.72

I solved the problem and once again, I got the wrong answer because I have the wrong sign for the acceleration. I attached the problem and books solution as well as my solution.
 

Attachments

  • Problem 12 72.pdf
    131.2 KB · Views: 323
  • #6
cipotilla said:
I solved the problem and once again, I got the wrong answer because I have the wrong sign for the acceleration.
I didn't go through your entire solution (I will if need be), but I note that r is defined as the distance between pulley and block A. So a positive acceleration for r means that the block accelerates to the left! I think that's at the root of the problem.

I note that you have a_A = - a_B/cos(theta); that tells me right there that you are (or should be) measuring a_A with positive values going to the left. Clearly the constraint is such that when block B moves down, block A moves to the right. (Tricky stuff, I know. But hang in there!)
 
  • #7
Ok, I solved the problem correctly, the signs make sense to me now. Going back to my original confusion as to why in P12.72 acceleration is not constant and in P12.126 it is. I've learned that in P12.126, acceleration is constant because the cables are all positioned at a constant angle, they hang vertically and remain vertical. Whereas in P12.72 that is not he case, the angle theta changes and so the tension in the cables either increases or decreases. Since the forces are not constant the the acceleration is not constant. (Its the sum of the forces that has to remain constant, right?) If in P12.126 the blocks were angled and that angle was changing with time, then we would not have a constant acceleration. Cables are tricky.
 

1. What is constant acceleration?

Constant acceleration refers to a steady rate of change in an object's velocity over time. This means that the object's speed is increasing or decreasing by the same amount in each unit of time.

2. How is constant acceleration calculated?

Constant acceleration can be calculated by dividing the change in an object's velocity by the time it took for that change to occur. This is represented by the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between constant acceleration and uniform acceleration?

Constant acceleration and uniform acceleration are often used interchangeably, but there is a subtle difference. Constant acceleration refers to a consistent change in velocity, while uniform acceleration refers to a consistent change in both velocity and direction. Essentially, constant acceleration can be either positive or negative, while uniform acceleration is always positive.

4. How does constant acceleration affect an object's motion?

Constant acceleration can have a significant effect on an object's motion. If the acceleration is positive, the object's speed will increase over time, and if the acceleration is negative, the object's speed will decrease. This means that the object will either be speeding up or slowing down, depending on the direction of the acceleration.

5. What is an example of constant acceleration in real life?

An example of constant acceleration in real life is a car traveling on a highway with cruise control engaged. The car's speed remains constant, but it is still accelerating due to the constant force of the engine. Another example is an object falling towards the ground due to the constant force of gravity, although air resistance may make the acceleration slightly non-uniform.

Similar threads

  • Introductory Physics Homework Help
Replies
29
Views
898
  • Introductory Physics Homework Help
Replies
6
Views
950
  • Introductory Physics Homework Help
Replies
13
Views
929
  • Introductory Physics Homework Help
Replies
2
Views
203
  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
883
  • Introductory Physics Homework Help
Replies
5
Views
361
  • Introductory Physics Homework Help
Replies
6
Views
744
  • Introductory Physics Homework Help
Replies
2
Views
663
Replies
3
Views
567
Back
Top