How Much Force Can Be Applied Before the Top Box Slips?

AI Thread Summary
To determine the maximum horizontal force that can be applied to the larger box without the smaller box slipping, the static friction force must be calculated using the coefficient of friction and the weight of the smaller box. The static friction force is given by the product of the coefficient of static friction (0.25) and the normal force, which is the weight of the smaller box (2.2 kg). The net force equations must be set up correctly, considering both static and kinetic friction. The user is seeking clarification on the logic behind the calculations and the correct approach to solving the problem. Understanding the relationship between the forces and the coefficients of friction is crucial for finding the solution.
Kakateo
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Homework Statement



A box with a mass of 2.2 kg sits on top of another box with a mass of 3.8 kg. The coefficient of friction between the two boxes is 0.25, and the coefficient or kinetic friction between the larger box and the horizontal surface is 0.32. Determine the largest horizontal force that can be applied to the larger box so that the smaller box does not slip off.

Homework Equations



Let fs be friction static
Let fs be friction kinetic

Fnet = Fa - Ffk
Fnet = Fa - Ffs

The Attempt at a Solution



I made Fnet for equation #2 0. I isolated each for force applied for each of the equations and set them equal to each other.

My equation before plugging numbers in for variables was this:

Ffk = Fnet + Ffs

I tried expanding the equation then solving for acceleration in which I got a negative value so I know I did something wrong.

I'd like to know the logic behind solving this equation and the process itselfThanks so much
 
Last edited:
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Hi Kakateo http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Determine the largest force that can be transferred (via friction) to the top box.
 
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