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A cup of coffee is on a table in an airplane flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates forward, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

2.9 $\mathrm{m} / \mathrm{s}^{2}$

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University of Michigan - Ann Arbor

Numerade Educator

Hope College

University of Winnipeg

as the airplane accelerates towards the right, say what happens is that the table. We'll try to slide below our cup of coffee, but luckily the frictional forces can save us. So what happens with the frictional force is that as the table tries to his light below or a cup of coffee to the right, the frictional force will try to bring the cup of coffee together with the table to the right. So the fictional force that appears is pointing to the right like this. Additionally, of course, there are the normal force on the weight force acting on that cup of coffee. So how can we calculate what is the maximum possible frictional force that can act on our cup of coffee? Well, the answer for this question is very simple. The maximum frictional force is the cause to the static frictional coefficient times the normal force that acts on our cup of coffee. In this question, the weight is precisely counterbalanced by the normal force. Therefore, the normal force is equal in magnitude to the weight force. So the fictional force is given by the static frictional coefficient, times the weight force, and this is equals to static frictional coefficient times the mass off that cup of coffee times, acceleration of gravity. Then we can relate to this frictional force with the plane acceleration using Newton's second law. So beautiful Second law tells us that the net force that acts on that cup of coffee is equals to the mass off the cup off coffee times, its acceleration. Then let's suppose that the cup off coffee acceleration is equals to the plane acceleration, because in the situation it isn't sliding over the table. So we have the static frictional coefficient times the mass off the cut off coffee times, the gravitational acceleration being ecause to the mess off that cup of coffee times its exploration. We have mass on both sides, so you can simplify, and we get the conclusion that the maximum possible acceleration that the plane can have in order for the cup of coffee not to slide over that table. Is it close to the StarTac frictional coefficient times acceleration of gravity As usual, we used the exploration off gravity as approximately 9.8 meters per second squared to get acceleration that is given by 0.3 times 9.8 then the maximum acceleration that the plane can have is approximately 2.9 meters per second squared, and this is our final answer.

Brazilian Center for Research in Physics