Friction between two blocks on a table

AI Thread Summary
In the first scenario, when the two blocks are at rest, the static friction from the table balances the tension in the rope, resulting in no static force exerted by the bottom block on the top block. In the second scenario, if the bottom block begins to slip due to a heavy mass descending, the kinetic friction force exerted by the bottom block on the top block acts in the direction opposing the slip. The discussion clarifies that friction always acts to counteract relative motion between the blocks. Additionally, the free body diagram for both blocks is reviewed, with corrections noted for the forces acting on the top block. Understanding these frictional forces is crucial for analyzing the system's dynamics.
burhan619
Messages
7
Reaction score
0
I have a question for two scenarios (see the diagram attached):

1) When the two blocks on the table are at rest, the static friction the table exerts on the bottom block is counteracting the tension in the rope, resulting in a net force of zero. Is there any static force being exerted by the bottom block on the top block? I don't think there should be any, since there is no horizontal force acting on the top block.

2) Suppose M was massive enough to start descending, overcoming the static friction threshold of the bottom block. Also suppose that the top block starts to slip away from the bottom block. Which way does the kinetic force the bottom block exerts on the top block point? And what's exerting the horizontal force it's counteracting?

Thanks in advance.
 

Attachments

  • friction illustration.jpg
    friction illustration.jpg
    1.9 KB · Views: 481
Physics news on Phys.org
burhan619 said:
I have a question for two scenarios (see the diagram attached):

1) When the two blocks on the table are at rest, the static friction the table exerts on the bottom block is counteracting the tension in the rope, resulting in a net force of zero. Is there any static force being exerted by the bottom block on the top block? I don't think there should be any, since there is no horizontal force acting on the top block.
You are correct.

2) Suppose M was massive enough to start descending, overcoming the static friction threshold of the bottom block. Also suppose that the top block starts to slip away from the bottom block. Which way does the kinetic force the bottom block exerts on the top block point?
Friction acts in a direction to oppose slipping. Which way would that be?
And what's exerting the horizontal force it's counteracting?
What do you mean? This is kinetic friction.
 
Is the free body diagram I've attached right?
 

Attachments

  • friction free body.jpg
    friction free body.jpg
    5.8 KB · Views: 722
here's a re-upload if that one's hard to see
 

Attachments

  • Untitled-1-01.jpg
    Untitled-1-01.jpg
    12.2 KB · Views: 485
burhan619 said:
Is the free body diagram I've attached right?
I guess the color coding is to tell which body the force acts on.

The forces on m1 look OK.

On m2, you forgot the kinetic friction from m1 and the normal force from m1. Otherwise, looks good.
 
That makes sense. Thanks.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top