Friction on Aluminum & Copper Blocks: Find Sum of Magnitudes

  • Thread starter Thread starter Ondra
  • Start date Start date
  • Tags Tags
    Friction
Click For Summary

Homework Help Overview

The problem involves a system of two blocks, one aluminum and one copper, connected by a string over a pulley, situated on a steel surface at an angle. The inquiry focuses on whether the blocks will move when released and, if not, the sum of the frictional forces acting on them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conditions for movement of the blocks, including the forces acting on them and the calculations related to friction. There are questions about the correctness of the calculations and the application of friction formulas.

Discussion Status

The discussion is ongoing, with participants examining the calculations made by the original poster and questioning the assumptions regarding the forces involved. Some guidance has been offered regarding the correct application of friction formulas, but no consensus has been reached on the final interpretation of the results.

Contextual Notes

Participants are considering the coefficients of friction provided and their sources, as well as the implications of the angle of the incline on the normal force calculations. There is a mention of previous threads and external resources that may provide additional context.

Ondra
Messages
1
Reaction score
0

Homework Statement


A 2.00-kg aluminum block and a 6.00-kg copper block are connected by a light string over a frictionless pulley. They sit on a steel surface as shown in Figure P5.64, where theta= 30.0°. When they are released from rest, will they start to move? If so, determine (a) their acceleration and (b) the tension in the string. If not, determine the sum of the magnitudes of the forces of friction acting on the blocks.

https://drive.google.com/file/d/0B4ksjSfetGUOeUdXX0tKODRWVjg/view?usp=sharing ---- IMAGE
fyzika.png


Homework Equations


1) I don't know how to calculate the last part: "If not, determine the sum of the magnitudes of the forces of friction acting on the blocks."

2) Is it so far correctly calculated?

The Attempt at a Solution


For the system to start to move when released, the force tending to move m2 down the incline, m g 2 sinθ, must exceed the maximum friction force which can retard the motion:

F = f(steel) + f(copper)

= µ1m1g + µ2m2g

= 0.61 x 2.0 x 9.8 + 0.53 x 6.0 cos(30) x 9.8 = 38.9 N

Force tending to cause the system to move = 6.0 x 9.8 x sin(30) = 29.4 N
Hence it will not start to move when released.
 
Last edited by a moderator:
Physics news on Phys.org
It just hit me... Did I already solved the: "If not, determine the sum of the magnitudes of the forces of friction acting on the blocks." Part? :D
 
Last edited by a moderator:
Ondra said:
µ2m2g
That is not correct for the block on the slope, but I see you used the right version in the next line.
Ondra said:
Did I already solved the: "If not, determine the sum of the magnitudes of the forces of friction acting on the blocks." Part?
Possibly. You calculated two forces. Which one are you suggesting as answer to part b?
Before answering, what exactly is the standard equation relating to static friction?
 
The formula for friction is F = coeffiicent of friction * Normal contact force. For M2, the normal contact force is not the weight. but you have corrected it lol. Where did you get the coefficients from? Online? I see its 0.61 between aluminium and steel.

Also, I know I posted in your earlier thread that got deleted, so please try to check out these two websites:

scroll down for mechanics:
http://www.cimt.org.uk/projects/mepres/alevel/alevel.htm
 
Bipolar Demon said:
The formula for friction is F = coeffiicent of friction * Normal contact force.
Not quite, and that is important here.
 
  • Like
Likes   Reactions: Logical Dog

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
16
Views
3K
Replies
2
Views
680
Replies
61
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
3K
Replies
18
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K