Determining max horizontal force using friction coefficient

In summary, the maximum horizontal force that can be applied to the lower block without the upper block slipping is 8.1N. This is determined by considering the frictional forces on both blocks and using Newton's second law to calculate the net external force on the system. Taking into account the coefficient of static friction between the two blocks and the coefficient of kinetic friction between the lower block and the table, the maximum force can be determined to be 8.1N.
  • #1
Monic
9
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.
 
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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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 teh lower block is accelerating both blocks).
 

Related to Determining max horizontal force using friction coefficient

1. What is the purpose of determining the max horizontal force using friction coefficient?

The purpose of determining the max horizontal force using friction coefficient is to understand the maximum amount of force that can be applied horizontally to an object without causing it to slide or move due to the force of friction.

2. How is friction coefficient related to the max horizontal force?

The friction coefficient is directly related to the max horizontal force. It is a measure of the amount of friction between two surfaces and is used to calculate the maximum force that can be applied without causing movement.

3. What factors affect the friction coefficient?

The friction coefficient can be affected by several factors including the type of surface, the roughness of the surface, the weight of the object, and the angle of the applied force.

4. How is the max horizontal force calculated using the friction coefficient?

The max horizontal force can be calculated by multiplying the friction coefficient by the normal force (the force exerted by the object on the surface). This will give the maximum amount of force that can be applied horizontally without causing movement.

5. Why is it important to determine the max horizontal force using friction coefficient?

Determining the max horizontal force using friction coefficient is important in understanding the limits of an object and ensuring its stability. It can also be used in designing structures and machinery to ensure they can withstand the forces that will be applied to them.

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