A very important application in mechanics is the inclined plane. The
following incline plane is assume to be frictionless. Inclined planes are
defined based on the angle with the horizontal, angle q
in the figure below. Given the angle q = 350
and the mass, m = 220Kg, of the box. How should this problem be
solved when calculating for the acceleration and the normal force?
Find the acceleration of the box:
None of the above.
and the normal force
None of the above.
In addition to follow the recommendations stated in
Application I, this problem should be initially solved by drawing the
actual vectors present in the problem
. These vectors are either forces or accelerations. In
the absence of friction, the vectors present are:
The weight of the box. This force is always
vertical directed to the center of the Earth. And,
The normal force on the box due to the
reaction of the surface of the inclined. This force is always
perpendicular to the reacting surface.
Since there is not any friction, nothing
prevent the box for sliding down the inclined. Therefore, there
is an acceleration along the direction of motion along the
Based on the vectors of the figure, a different
coordinate system should be selected. The standard horizontal x-axis
and vertical y-axis are not the most convenient one for this
situation. Notice that there are two vectors that are already
perpendicular to each other, the acceleration and the normal force.
Thus, the coordinate system should be selected along these directions
. In this coordinate system, the only vector that need to
be broken down into components is the weight of the box.
A very important step in the solution of this problem
is to find out where between the vectors and the coordinate system is
the angle q replicated
. Notice that the extension of the y-axis to the
horizontal, the horizontal, and the inclined plane itself form a right
angle triangle with known angles, 900 , 900 -
q, and q . A
second triangle is formed by the same extension of the y-axis to
the horizontal, the horizontal and the vertical extended from the vector
representing the weight of the object
. The x-component of the weight correspond to the
opposite side with respect to the angle q
. After finding the y-component of the weight
(adjacent side with respect to the angle q),
the diagram associated with this problem is completed
. Finally, the FBD of the problem can be obtained
. Now, from the diagram, the equations involving the
unknowns, a and FN, can be obtained. This
equations are obtained by analyzing each one of the components.
From the diagram (Last figure of the previous sequence), the net force
in the x direction is
Applying Newton's second law to this component,
The first equation is obtained equating the two
previous expressions for the normal force,
From where the algebraic form of the acceleration is
And the net force in the y direction is
Applying Newton's second law to the y-component,
Notices that there is not motion along the y-direction.
Therefore, the acceleration along this direction is zero. Equating the
two previous relations, the second equation is obtained,
The numerical solution is
Thus, the acceleration is
normal force is