Acup 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 0.180 . 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.

We know that the formula for static friction is:

Fs = μ N

where,

μ is the coefficient of static friction = 0.180

N is the normal force = weight of the cup = m g

We also know that from the Newton’s second law of motion, the formula for net force is:

Fnet = m a

where m is the mass of the cup and a is the acceleration of the cup

To prevent the cup from sliding backward, the static friction Fs must cancel out the net Force Fnet, that is:

Fnet – Fs = 0

Therefore:

m a – μ N = 0

m a = μ N

m a = μ (m g)

Cancelling m:

a = μ g

a = 0.180 (9.81 m / s^2)

a = 1.7658 m / s^2

Therefore the plane can only have an acceleration of about 1.77 m/s^2 before the cup starts to slide.

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