What is the maximum force of static friction for the block?

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving static friction, a block, and a hanging bucket. The original poster presents a scenario where a block is subjected to a force via a string connected to a bucket, with the goal of determining the maximum force of static friction and the number of washers that can be added to the bucket without moving the block.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between static friction and the forces acting on the block and the hanging mass. There are attempts to calculate the maximum static friction force and the implications of adding washers to the bucket. Questions arise regarding the assumptions made about acceleration and the forces involved.

Discussion Status

Some participants have provided guidance on understanding the forces at play, particularly the relationship between tension and static friction. There is ongoing exploration of the calculations and assumptions, with no explicit consensus reached on the correct approach to the problem.

Contextual Notes

Participants note that the coefficient of static friction is given, and there is a misunderstanding regarding the forces acting on the hanging mass and the block. The discussion reflects confusion about the values used in calculations and the conditions under which the block remains stationary.

oMovements
Messages
28
Reaction score
0

Homework Statement


A string is tied to a 4.4 kg block and 120g hanging bucket. Students add 20g washers one at a time to the bucket. The student are unaware that the coefficient of static friction for the block on the table is 0.42.
A) what is the maximum force of static friction for the block?
B) how many washers can the students add to the bucket without moving the block?

2. The attempt at a solution
a)FssFn
= (0.42)(4.4x9.8)
Fs= 18.1N

b) Ft-Fs=ma
Ft-18.1=4.4a
Ft=4.4a+18.1
Ft=1.6N

Ft-Fg=m(-a)
4.4a+18.1=1.176=0.12a
4.52a=16.924
a=3.74m/s/s

I don't know what the next step after this would be. Right now I know that the hanging mass has a tension of 1.6N so another 16.5N can be added to the tension force to reach the maximum static force.

I also tried this:
Fg=mg
16.5=(0.12+0.02x)(9.8)
16.5=1.176+0.196x
15.324=0.196x
x=78 washers

The correct answer is 86 washers, but I don't know how to get to this answer.
 
Last edited:
Physics news on Phys.org
oMovements said:
Fg=mg
16.5[/color]=(0.12+0.02x)(9.8) ✗[/size][/color]
Should be 18.1=(0.12+0.02x)(9.8)

The friction is overcome by the (bucket + washers). The string just conveys the force, the frictionless pulley changes the direction of the force. Acceleration is 0 until the friction is overcome.

The correct answer is 86 washers, but I don't know how to get to this answer.
You should be able to do it now. :smile:
 
oMovements said:
b) Ft-Fs=ma
Ft-18.1=4.4a
Ft=4.4a+18.1
Ft=1.6N

The thing that confuses me about your solution is this bit in red here. How did you come up with this number of 1.6 N? It seems like you assumed a value for 'a' (that was wrong).

The solution is much simpler than you are making it. You are correct that the only two horizontal forces acting on the block are static friction, and the force due to the tension in the rope. If the tension force is *just about* to overcome static friction, what must be true about the value of Ft as compared to Fs, and what is the value of 'a' in that case? (EDIT: NascentOxygen already gave you the answer to that last question about 'a'). The answer to these questions tells you all you need to know.
 
I still do not understand why Fg=18.1. I know that the hanging mass must have a maximum net force of 18.1N without the block moving. However, shouldn't Ft-Fg=18.1 because those are the two forces acting on the hanging mass. Or is Fg=18.1 because the acceleration is zero, therefore so is the net force and as a result Ft= -Fg and they both cancel each other out which doesn't cause the object to move. Is that correct?
 
oMovements said:
Or is Fg=18.1 because the acceleration is zero, therefore so is the net force and as a result Ft= -Fg and they both cancel each other out which doesn't cause the object to move. Is that correct?

Yes.
 

Similar threads

Replies
61
Views
3K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
19
Views
5K
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
12K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K