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 0.420 . 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? Use g = 9.81 m/s2.
To find the maximum acceleration that the plane can have without the cup sliding backward, we need to calculate the maximum static friction force that can act on the cup.
The maximum static friction force can be calculated using the formula:
F_fric_max = u_static * N
where F_fric_max is the maximum static friction force, u_static is the coefficient of static friction, and N is the normal force.
In this case, the normal force N is equal to the weight of the cup, which can be calculated using the formula:
N = m * g
where m is the mass of the cup and g is the acceleration due to gravity.
Since the cup is on a table in an airplane flying at a constant altitude, the weight of the cup remains the same as its mass times the acceleration due to gravity.
Now, we can substitute the values given in the question:
u_static = 0.420 (coefficient of static friction)
g = 9.81 m/s^2 (acceleration due to gravity)
To find the maximum acceleration, we need to determine the maximum static friction force that occurs just before the cup begins sliding. In this case, the maximum static friction force will be equal to the force of inertia acting on the cup in the backward direction due to the acceleration of the plane.
Therefore, we can set up the equation:
F_fric_max = m * a_max
where F_fric_max is the maximum static friction force, m is the mass of the cup, and a_max is the maximum acceleration of the plane.
Since F_fric_max is also equal to u_static * N, we can equate the two equations:
u_static * N = m * a_max
Substituting the values of N and rearranging the equation, we get:
u_static * m * g = m * a_max
Simplifying further, we find:
a_max = u_static * g
Now we can substitute the values of u_static and g:
a_max = 0.420 * 9.81
Calculating the result, we find:
a_max ≈ 4.11042 m/s^2
Therefore, the maximum acceleration that the plane can have without the cup sliding backward on the table is approximately 4.11042 m/s^2.