Why Is the Direction of Friction on Block A to the Right?

In summary: I think my professor would disagree because the block doesn't go very far.Wow that makes so much sense now, the block slides backwards. Thank you so much. However... is it really possible? I think my professor would disagree because the block doesn't go very far.
  • #1
theBEAST
364
0

Homework Statement


I attached a picture of the problem and I was wondering why the direction of the friction on block A is to the right. Shouldn't it be to the left because friction always opposes the direction of the acceleration (which is in this case to the right for block A)?

Thanks!
 

Attachments

  • Capture.PNG
    Capture.PNG
    11.3 KB · Views: 521
Physics news on Phys.org
  • #2
no, the friction is to the right. Your definition of friction 'opposing the direction of acceleration' is not always true. This is probably why you're having trouble. If you think about what friction does, then you can come up with a better rule for deciding the direction friction will act. (just think intuitively about it)
 
  • #3
BruceW said:
no, the friction is to the right. Your definition of friction 'opposing the direction of acceleration' is not always true. This is probably why you're having trouble. If you think about what friction does, then you can come up with a better rule for deciding the direction friction will act. (just think intuitively about it)

Thanks, also is there a good way to determine whether or not the block will slide? I recall my professor teaching us one method that says that if the calculated acceleration is greater than zero when you calculate your numbers with the static friction the block will slide. However in this question I tried that and even when i plug in a coefficient of friction = 1000 the acceleration is still positive. According to my professor this means that the block will still slide. But intuitively that is impossible with such a high coefficient of friction.


Here is my work to show you what I mean, focus on the red text.

http://dl.dropbox.com/u/64325990/Photo%202012-03-10%2012%2038%2000%20AM.jpg
 
Last edited by a moderator:
  • #4
That method your professor taught you is not correct for this type of question. I'm guessing that method was used for a question about a block on an incline? In any case, it is not right for this question.

Looking at your jpg, you've done pretty well so far. But it looks like you've used 44 as the mass of the cart, even though the question says the cart has 30kg mass. Anyway, apart from that it is good so far. So you've got the acceleration of the box and the acceleration of the cart. What is the next step to showing that they slide past each other?
 
  • #5
BruceW said:
That method your professor taught you is not correct for this type of question. I'm guessing that method was used for a question about a block on an incline? In any case, it is not right for this question.

Looking at your jpg, you've done pretty well so far. But it looks like you've used 44 as the mass of the cart, even though the question says the cart has 30kg mass. Anyway, apart from that it is good so far. So you've got the acceleration of the box and the acceleration of the cart. What is the next step to showing that they slide past each other?

WOW you are right on! It was a block on an inclined plane. Do you know why the method only works for those kinds of problems?

As for my question, the textbook has 44 kg mass but for my homework mine is 30 kg mass. So I am guessing I am suppose to compare the acceleration of the box and cart? But this is the acceleration using static friction? Using this I found that the acceleration of the cart is 3.21 and the block is 2.94. Hmmm... I am not sure what to do next.
 
  • #6
theBEAST said:
As for my question, the textbook has 44 kg mass but for my homework mine is 30 kg mass. So I am guessing I am suppose to compare the acceleration of the box and cart? But this is the acceleration using static friction? Using this I found that the acceleration of the cart is 3.21 and the block is 2.94. Hmmm... I am not sure what to do next.
Yes, it should be calculated using the static coefficient of friction (since we are trying to see if it is possible for the block to not slip). Allright, so you've got the acceleration of the block and the cart. You can see that the acceleration of the cart is greater than the acceleration of the block. So if you imagine the situation, what will happen?
 
  • #7
BruceW said:
Yes, it should be calculated using the static coefficient of friction (since we are trying to see if it is possible for the block to not slip). Allright, so you've got the acceleration of the block and the cart. You can see that the acceleration of the cart is greater than the acceleration of the block. So if you imagine the situation, what will happen?

Wow that makes so much sense now, the block slides backwards. Thank you so much. However I would really like to know why the method for objects on an inclined planes cannot be applied to this question.
 
  • #8
Its the same principle. The inclined plane has zero acceleration, so what must be the acceleration for the block so that the block does not slip relative to the plane?

And yes, for this question, the block slides backwards. The next part of the question is to work out the time it takes the block to move 1.5m relative to the cart.
 
  • #9
BruceW said:
Its the same principle. The inclined plane has zero acceleration, so what must be the acceleration for the block so that the block does not slip relative to the plane?

And yes, for this question, the block slides backwards. The next part of the question is to work out the time it takes the block to move 1.5m relative to the cart.

Thanks for your help I got the right answer :D

However I would still like to clear up some confusion regarding the block on an inclined plane question. The tricky part to this problem is figuring out which way the acceleration is directed and thus which way the friction is directed. Here is the problem:

http://dl.dropbox.com/u/64325990/Inclined%20Plane%20Question.jpg

The diagram I drew in this picture uses the correct acceleration which according to my professor you have to guess and see if the answers come out right.


So I decided I would attempt this problem with the acceleration in the wrong direction and see what I get. Here is my attempt, the solution above the red line is with the acceleration in the incorrect direction and below the red line is the correct direction with the correct answer:

http://dl.dropbox.com/u/64325990/Inclined%20Plane%20Solution.jpg

NOW what I don't understand is how do you know if you chose the wrong direction for the acceleration. In my case I got a negative acceleration using the static coefficient of friction. HOWEVER, our professor said that if the acceleration is negative it just tells us that the block will not slide. So say on an exam if I wrote out the solution above the red line in the wrong direction it should be still correct? It just implies that the block does not slide.
 
Last edited by a moderator:
  • #10
Does anyone know how to answer my previous question? :S
 

Related to Why Is the Direction of Friction on Block A to the Right?

1. What is a free body diagram?

A free body diagram is a visual representation that shows the forces acting on an object. It is used in physics and engineering to analyze the motion and equilibrium of an object.

2. Why is a free body diagram important?

A free body diagram is important because it helps us understand the forces acting on an object and how they affect its motion. It allows us to accurately analyze the forces and make predictions about the object's behavior.

3. How do I draw a free body diagram?

To draw a free body diagram, first identify the object you want to analyze. Then, draw a dot to represent the object and label it with the object's name. Next, draw arrows to represent the forces acting on the object, making sure to label each force with its name and direction. Finally, make sure all forces are drawn to scale and in the correct direction.

4. What are the key components of a free body diagram?

The key components of a free body diagram are the object, forces acting on the object, and the direction and magnitude of each force. It is also important to label each force and make sure they are drawn to scale.

5. How can a free body diagram be useful in problem-solving?

A free body diagram can be useful in problem-solving by helping us identify and understand the forces affecting an object. By accurately representing these forces, we can use equations and principles of physics to solve for unknown variables and predict the behavior of the object.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
588
  • Introductory Physics Homework Help
Replies
10
Views
384
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
866
  • Introductory Physics Homework Help
Replies
11
Views
834
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
998
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
30
Views
2K
Back
Top