A box, placed on an inclined plane does not slide. A slight push, however, cause
ID: 1502874 • Letter: A
Question
A box, placed on an inclined plane does not slide. A slight push, however, causes the box to slide down the plane at an increasing speed. Which of the following equations shows why the box is in an unstable equilibrium?
a. Fnet = 0
To solve the problem, you must write a clear and convincing justification for accepting one of the possible answer choices as the best or most appropriate and for rejecting each of the other choices as less appropriate. Your reasons for rejecting choices are just as important as your reasons for choosing the best answer.
Explanation / Answer
Hi,
In this case we have a box on a plane with an angle over the horiziontal. The surfaces of the plane and the box are such that cause friction between them. This means that over the box there are three forces exerting influence: the weight, the normal force and the friction.
Choosing the right answer
If we construct a free body diagram of the box with the x axis parallel to the plane and the y axis perpendicular to it then we have the following:
x axis: fs = mg sin()
y axis: FN = mg cos()
Note: it may be an unstable equilibrium, but it is an equilibrium nonetheless, so there is no acceleration at any axis.
The friction force and the normal force are related by: fs < u*FN ;where u is the static friction coefficient.
However, there's a moment where the friction force is about to be surpass and in that instant the body (the box in this case) is in an unstable equilibrium, mathematically this is:
fs = u * FN ::::::::::: u = fs / FN
The previous statement implies that there is a critical angle in which the movement will begin and that value will be:
u = mg sin () / mg cos () :::::::: u = tan() ; where is the value of the critical angle.
To sum up, two ways to express that this system is in an unstable equilibrium are:
u = fs / FN or =
So we have to conclude that the option b is the right one.
What happens with the others?
Letter a. This one only says that the net force is equal to cero. This means that the acceleration of the box is cero, but this could imply two different situations:
1. The box is not moving at all.
2. The box is moving but with constant speed.
As there is a possibility of movement and the problem says that the box is at rest, this cannot be the answer.
Letter c. This one is telling us that the torque about a certain axis is equal to cero. This implies that the object is not rotating or it is rotating but at a constant pace.
As this has nothing to do with the unstability of the box and only assure us the condition of mechanical equilibrium, this cannot be the answer.
Letter d. This one says that the kinetic coefficient of friction is smaller than the static one, but this is always true so in fact this letter is not giving us anything.
Letter e. This one is similar to Letter d, because if you put the axis as were put before, the normal force will always be equal to that component of the weight. So this one cannot be the answer.
I hope it helps.
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