Determining max horizontal force using friction coefficient

Click For Summary

Homework Help Overview

The problem involves determining the maximum horizontal force that can be applied to a lower block without causing an upper block to slip, given specific coefficients of friction and masses of the blocks. The subject area includes concepts of friction, Newton's laws, and dynamics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the blocks, including frictional forces and how they relate to acceleration. There are attempts to calculate frictional forces and subsequent accelerations, with some participants questioning the assumptions made about these calculations.

Discussion Status

The discussion is ongoing, with participants providing guidance on correcting calculations related to frictional forces and considering all forces acting on the blocks. There is recognition of the need to clarify the relationship between the forces and the resulting accelerations.

Contextual Notes

Participants note potential errors in calculations, such as not accounting for gravitational effects on friction and the need to use the correct normal forces. There is also mention of homework constraints that may limit the approach to the problem.

Monic
Messages
9
Reaction score
0

Homework Statement


A 0.5 kg wooden block is placed on top of a 1.0kg block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.2. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping? The correct answer on the sheet is 8.1N

Homework Equations


aΔt=v2-v1
d= 1/2(v1 + v2)∆t
d=v1Δt + 1/2aΔt^2
v2^2=v1^1 + 2ad
F=ma
μ=Ff/Fn

The Attempt at a Solution



(9.8)(0.2)=Ff of bottom 1kg box
1.96N=Ff

(0.5)(0.35)=Ff for top 0.5 kg box
0.175N=Ff

Unsure of the next step. I assumed the force would be equal to the frictional force of the bottom box but it is not.
 
Physics news on Phys.org
There is only one force acting on the upper block. You could try considering what the maximum value of that force could be, and hence what the maximum possible acceleration of the upper block is going to be.
 
The maxiumum value of force is the same as the Ff (0.175 as shown above) If I calculate accelertion 0.175N/0.5kg=a i get 0.35m/s^2

If I put that acceleration to calculate the max force of the other block (1kg)(0.35m/s2)=F i get 0.35N again...which is wrong
 
You first should correct your value for the friction force on the top block...you forgot to multiply by g.
Then for the friction force on the bottom block from the table, you must use the normal force, not the weight. And when applying Newton 2 to the system, you must use the net external force.
 
Monic said:
If I put that acceleration to calculate the max force of the other block (1kg)(0.35m/s2)=F i get 0.35N again...which is wrong

Have you included all the forces acting on the lower block to get it to accelerate at particular rate? There's the frictional force from the table to consider... And if friction is accelerating the upper block as well, there's an equal and opposite force to consider (or you could just consider that as long as the upper block isn't sliding, the force applied to the lower block is accelerating both blocks).
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
13
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
12K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
18
Views
3K
Replies
1
Views
3K