Determining max horizontal force using friction coefficient

AI Thread Summary
To determine the maximum horizontal force that can be applied to a lower block without the upper block slipping, the coefficients of static and kinetic friction are crucial. The static friction force between the blocks is calculated as 0.175N, while the kinetic friction force between the lower block and the table is 1.96N. The maximum applied force must account for both blocks, as the upper block's friction affects the overall system. Correctly applying Newton's second law and considering all forces acting on the lower block is essential for accurate calculations. The final answer for the maximum horizontal force is 8.1N, aligning with the provided solution.
Monic
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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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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 teh lower block is accelerating both blocks).
 
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