A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.37. Suddenly, the plane accelerates, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?
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To determine the maximum acceleration the plane can have without the cup sliding backward on the table, we need to consider the forces acting on the cup.
1. Weight (W): The force due to gravity acting vertically downward on the cup.
2. Normal force (N): The force exerted by the table on the cup, perpendicular to the table's surface.
3. Frictional force (F): The force opposing the motion of the cup when it tries to slide backward.
Since the cup is not sliding, the maximum static friction force is equal to the force exerted by the table on the cup:
F_max = μ_s * N
= 0.37 * N
The normal force (N) is equal to the weight of the cup:
N = W = mg
where:
m = mass of the cup
g = acceleration due to gravity
Now, let's consider the forces involved during the acceleration of the plane:
1. Weight (W): Same as before.
2. Normal force (N): Same as before.
3. Horizontal force (F_h): The force that accelerates the cup along with the plane.
Since the plane is accelerating horizontally and the altitude remains constant, the net force in the horizontal direction is provided by the horizontal force:
F_h = ma
where:
m = mass of the cup
a = acceleration of the plane
To prevent the cup from sliding backward, the maximum frictional force must be equal to the horizontal force:
F_max = F_h
0.37 * N = ma
Since N = W, we can substitute:
0.37 * mg = ma
Simplifying the equation and solving for the acceleration (a), we have:
a = (0.37 * g)
Therefore, the maximum acceleration the plane can have without the cup sliding backward on the table is 0.37 times the acceleration due to gravity (g).
To determine the maximum acceleration that the plane can have without the cup sliding backward on the table, we need to consider the forces acting on the cup.
First, let's examine the forces acting on the cup when it is at rest on the table. The force of gravity pulls the cup downward, while the force of static friction acts in the opposite direction, opposing the motion of the cup. These two forces are equal in magnitude and opposite in direction, preventing the cup from sliding.
Now, when the plane accelerates forward, an additional force is applied to the cup in the forward direction. This force is due to the inertia of the cup, resisting the change in its state of motion. The static friction force needs to increase to match this additional force and prevent the cup from sliding backward.
The maximum static friction force (F_max) that can be exerted between the cup and the table is given by the equation:
F_max = coefficient of static friction (μ) × normal force (F_n)
The normal force (F_n) is the force exerted by the table on the cup, which is equal in magnitude and opposite in direction to the force of gravity on the cup.
Now, since the altitude of the plane remains constant, the force of gravity will still act vertically downward. Therefore, the normal force (F_n) will also be vertical and equal in magnitude to the force of gravity.
To find the maximum acceleration, we need to set up an equation equating the maximum static friction force (F_max) to the additional force exerted on the cup due to the acceleration. This additional force is given by:
F_additional = mass of the cup (m) × acceleration (a)
Setting F_max = F_additional, we have:
μ × F_n = m × a
Since F_n = weight of the cup = mass of the cup (m) × acceleration due to gravity (g), we can rewrite the equation as:
μ × m × g = m × a
Simplifying, we find:
a = μ × g
Now, we can substitute the given coefficient of static friction (μ = 0.37) and the acceleration due to gravity (g ≈ 9.8 m/s^2) into the equation to find the maximum acceleration:
a = (0.37) × (9.8 m/s^2)
The maximum acceleration that the plane can have without the cup sliding backward on the table is approximately:
a ≈ 3.626 m/s^2