# The disappearing force

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### Luigi Fortunati

May 14, 2022, 2:05:32 PMMay 14
to
In my animation
https://www.geogebra.org/m/qc9pfvva
there is the force of gravity <mg> (in black), the component parallel
to the plane (in blue) and the other component perpendicular to the
plane (in red).

Starting the animation, as the inclination of the plane increases, the
vertical force of gravity (in black) remains unchanged and the force
perpendicular to the plane (in red) decreases to zero.

Is it correct to say that as the inclination of the plane increases,
the red force disappears but the black force does not disappear at all?

[[Mod. note -- For anyone unable to view the animation, it shows a block
on an inclined plane (at an angle $\alpha$ to the horizontal), with the
block's weight $mg$ (shown in black) resolved into components
$mg \cos \alpha$ perpendicular to the plane (shown in red)
and $mg \sin \alpha$ parallel to the plane (shown in blue).

To answer the author's question: yes, the black force (the block's weight
$mg) is unchanged as the inclination of the increases, but the red force (the component$mg \sin \alpha\$ perpendicular to the plane) decreases.
When the inclination reaches 90 degrees then the red force is zero.
-- jt]]

### Luigi Fortunati

May 16, 2022, 11:47:12 AMMay 16
to
https://www.geogebra.org/m/jkvhjmjv
the "final position" button and the constraining force (green).

In the direction perpendicular to the plane, the red force and the
green force (equal and opposite) are reduced more and more until they
disappear at 90 degrees of the alpha angle.

Conversely, the blue force increases more and more until it becomes
equal to <mg> when the alpha angle reaches 90 degrees.

Is it correct to say that, if the block slides without friction along
the plane, its acceleration will increase as the alpha angle increases,
until it reaches its maximum value <g> at 90 degrees?