A classical mechanics problem by Kelvin Hong

Classical Mechanics Level 3

There is a unit tetrahedron $ABCD$ with face $BCD$ on the ground. Edge $CD$ is fixed to the ground so that vertex $B$ can be lifted up. Now, James wants to exert a force horizontally on point $A$ and turn over the tetrahedron, making vertex $B$ the new apex with face $ACD$ on the ground.

If the mass of the tetrahedron is $m$ , and the minimum force that is necessary to pull it can be expressed as $\frac{mg}{a\sqrt{b}}$ , where $b$ is a square-free number, then submit $a+b$ .

Details and Assumptions:

$g$ is the gravitational acceleration.
A unit tetrahedron is a tetrahedron with all six sides of length 1.

The answer is 4.

1 solution

Kelvin Hong
Aug 31, 2017

In the picture, $E$ is center of the triangle $BCD$ .

$x=EF= \frac{\sqrt3}{6}$

Then, if we connect $AE$ , it will be the high $h$ of the tetrahedron, and which is perpendicular to the ground.

It is not hard to find that $BE=\frac{\sqrt3}{3}$ , $AB=1$ ,

$h^2 + BE^2 = AB^2$

$h=\frac{\sqrt6}{3}$

To turning the tetrahedron, $torque_F \geq torque_{mg}$

$Fh \geq mgx$

Applying all of the numbers,

$F \geq \frac{mg}{2\sqrt2}$

$F_{min} = \boxed{\frac{mg}{2\sqrt2}}$

A classical mechanics problem by Kelvin Hong

The answer is 4.

1 solution

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