2 Blocks, 1 Cord [Acceleration, Tension]

In summary: F_{net} = ma = mg\sin\theta - TIf the tension is greater than mg\sin\theta, then it will accelerate in that direction. If the tension is less than mg\sin\theta, then it will accelerate in the opposite direction.It is correct so far, the blocks will slide to the right as the acceleration of the 50kg block is greater than the 100kg block and since they are connected by the same cord, they will move in the same direction. As for the values, you seem to have mixed up the masses. The 100kg block should have an acceleration of 5.87 m/s^2 while the 50kg block should have an
  • #1
Heat
273
0

Homework Statement



Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes:
YF-05-70.jpg

Which way will the system move when the blocks are released from rest? Right or left?
What is the acceleration of the blocks?
What is the tension in the cord?

Homework Equations



F=ma
w=ma

The Attempt at a Solution



First off all, it should move to the left, as there is more weight there.

We know that there is a force of gravity acting on each of the blocks straight downwards.

And the pulling back of the block 100kg is not much, thus allowing it to slide to the left.
For the 50kg block, the tension is greater than the normal force and the gravity force which allows it to move.

The acceleration of the system is the total sum of the forces divided by the sum of the masses.

The tension of the cord is the difference between the forces at the end of each.

If my description is correct, how do I go into the mathematical part.?
 
Physics news on Phys.org
  • #3
Heat said:
First off all, it should move to the left, as there is more weight there.
What matters is the component of gravity down the incline.

We know that there is a force of gravity acting on each of the blocks straight downwards.

And the pulling back of the block 100kg is not much, thus allowing it to slide to the left.
For the 50kg block, the tension is greater than the normal force and the gravity force which allows it to move.
Tension and normal force are perpendicular to each other. What allows it to accelerate is a nonzero net force.

The acceleration of the system is the total sum of the forces divided by the sum of the masses.
True, if you do it right. But I advise against taking such shortcuts at first.

The tension of the cord is the difference between the forces at the end of each.
What makes you think that?

If my description is correct, how do I go into the mathematical part.?
Rather than trying to do it in your head, attack it systematically:

Examine the forces parallel to the incline acting on each mass. Apply Newton's 2nd law to each.
 
  • #4
ok I got the acceleration, with the tip you provided snoop. :D,

now I will try tension. :)
 
  • #5
Heat said:
ok I got the acceleration, with the tip you provided snoop. :D,

now I will try tension. :)

You mean the shortcut? But I agree with Doc Al... I advise doing this from the basics...
 
  • #6
:O

Actually that was the same equation I found in my textbook. I guess I should do it from the basics, so I can understand this more, than just plug in.
 
  • #7
Heat said:
:O

Actually that was the same equation I found in my textbook. I guess I should do it from the basics, so I can understand this more, than just plug in.

Yeah, I agree... the reason I brought that formula up is because we can get much more complex variations on the atwood machine... and in those cases it might not be practical to go through it step by step...

But definitely in this problem, and while you're still learning about the atwood machine, you should do the problem from the basics.

That thread is still very helpful. Dick gives good suggestions on how to solve the problem from the basics.
 
  • #8
I still do not seem to understand how to get the x and y components here. :(

Sorry if you find this question annoying.

I tried shifting the x and y graph so that the tension of the 100kg block in on the x axis.
 
  • #9
Heat said:
I still do not seem to understand how to get the x and y components here. :(

Sorry if you find this question annoying.

I tried shifting the x and y graph so that the tension of the 100kg block in on the x axis.

I recommend reviewing this website if you're having trouble with the incline stuff:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html [Broken]

then try to apply that stuff here...
 
Last edited by a moderator:
  • #10
ok I was able to determine the acceleration of both objects.

The one that has 100 kg has acceleration of 2.94 and the 50kg has 5.87.

This must mean that the 50kg objects is going down and that the 100kg object is going up.

So the blocks should slide to the right.

Is this correct thus far?

PS, I know realized that the values may be incorrect, I will keep on analyzing, but is the direction of sliding correct?
 
Last edited:
  • #11
Heat said:
The one that has 100 kg has acceleration of 2.94 and the 50kg has 5.87.
The blocks are attached by the cord. How can the magnitude of their accelerations differ?

How did you arrive at this?

Did you analyze the forces acting on each mass? Apply Newton's 2nd law?
 
  • #12
Yes, I realized that it was wrong, since like you mentioned they are attached by a cord and should have the same acceleration magnitude wise.

I got this by drawing them into different figures, and got the x and y component.

I found that the x component of the 100kg object is 100(9.8) sin 30 and y component it was 100(9.8)cos30.

For the 50kg object, the x of 100(9.8)sin 53.1 and y of 100(9.8)cos 53.1.

This followed as well as I could understand the explanations listed in the similar scenario by Snoopy of getting the sum of Fx which gave me 293.69 and after applying the F=ma...I got different accelerations. This should not be the case.

The forces acting on each mass is as follows, object 100kg has a gravity and normal force equal in size, and a tension to the right. object 50kg has gravity and normal force equal in size, and a tension to the left.
 
  • #13
The normal force doesn't equal gravity (which is mg) but it equals the y-component of gravity ([itex]mg\cos\theta[/itex]).

Analyze each mass separately. (You'll connect them later by what you say about how their accelerations must be related.)

We know that the masses only move in the x-direction, and that the y-direction forces balance. So, what forces in the x-direction act on m_1? On m_2?

You can figure out the direction that they will accelerate by comparing the x-component of gravity on each mass. Which ever one pulls more, that's the direction of the acceleration.

But you don't have to know the acceleration--or its direction--it will come out of the equations. If you want, just guess the direction and solve for the acceleration. You'll know by the sign of your answer if you picked the correct direction. (A negative sign means it goes opposite to what you thought.)
 
  • #14
Ok, I redrew it and analyzed the object of 100kg.

I determined that it looks something like this.

300d3wl.jpg


If this is correct, then the normal force should be 848.71 right?

"So, what forces in the x-direction act on m_1?"

The forces on x direction would be 490.

I just did another one for the 50kg object, I found that the x component is 391.85 and the y is 294.21.

"You can figure out the direction that they will accelerate by comparing the x-component of gravity on each mass. Which ever one pulls more, that's the direction of the acceleration."

The one pulling more is the 100kg object. So it would slide to the left. (This is of course, if I determined the components correctly)
 
Last edited:
  • #15
It looks like you've got it. The x-component of gravity for each mass is [itex]mg\sin\theta[/itex].

So the system accelerates to the left. Call the magnitude of the acceleration "a" and the tension "T".

Let's deal with m_1 (100 kg) first: You have gravity acting down the incline, so that force should be called [itex]-m_1g\sin\theta_1[/itex]. (The minus sign indicates "to the left and down"; a plus sign means "to the right and up".) The tension is +T. So the total force is: T - [itex]m_1g\sin\theta_1[/itex]. Since the acceleration is down, we'd better call it -a.

Combining all of this into Newton's 2nd law, we get:
[tex]T - m_1g\sin\theta_1 = m_1(-a) = -m_1a[/tex]

Now do the same kind of analysis for m_2.

You'll end up with two equations. Solve them together and you'll get the acceleration and the tension.
 
  • #16
ok,

for m_2

we have gravity acting down on the incline, so the force will also be called -m_2gsin(angle). I would say the tension is also +T (don't the tension have to be the same as they are the same cord). So far it is T - m_2gsin(angle). But now the acceleration is up, so it would be +a.

When put together it would be.

T - m_2 g sin (angle) = m_2 (a) = m_2.Solving for acceleration:

T-m_1gsin(angle) = -m_1a
T - m_2 g sin (angle) = m_2a.

Ok, after pluggin in and solving for a and t

I got that a = .654 and that t = 424.57

hopefully this seems right. :)

well I know that acceleration is right, because, I took the shortcut earlier and got the same answer, but now I did it with the basics and it feels great achieving this. Now I hope my T is right. :D
 
Last edited:
  • #17
Very good! Once you get the hang of the "from the basics" method, you'll be able to solve all sorts of problems without having to memorize any special results. It's the way to go.

Here's your assignment. Study the relevant examples on this page: Standard Newton's Laws Problems
 
  • #18
Doc Al, thank you for taking time to teach me the basics.
I will study the examples on the page you provided, so I can get better at this and one day master physics.:smile:
 

1. What is the concept behind "2 Blocks, 1 Cord [Acceleration, Tension]"?

The concept behind "2 Blocks, 1 Cord [Acceleration, Tension]" is to understand the relationship between acceleration and tension in a system with two blocks connected by a cord. This system can be used to demonstrate the concepts of Newton's laws of motion.

2. How does the acceleration of one block affect the tension in the cord?

The acceleration of one block will directly affect the tension in the cord. When one block accelerates, it pulls on the cord, causing tension. The higher the acceleration, the greater the tension in the cord.

3. Can the tension in the cord be greater than the weight of the blocks?

Yes, the tension in the cord can be greater than the weight of the blocks. This is because the tension in the cord is determined by the forces acting on the blocks, which may include external forces such as friction or applied forces.

4. What factors can affect the acceleration and tension in the system?

The factors that can affect the acceleration and tension in the system include the mass of the blocks, the force applied to the blocks, and any external forces acting on the blocks, such as friction or air resistance.

5. How can "2 Blocks, 1 Cord [Acceleration, Tension]" be used in real-life applications?

"2 Blocks, 1 Cord [Acceleration, Tension]" can be used to model various real-life scenarios, such as the movement of elevators, pulley systems, and even the motion of objects on inclined planes. It can also be used in engineering and physics experiments to study the effects of forces on a system.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
858
Replies
6
Views
202
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
167
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top