# A plane is raise 8 inches in the front and lowered 8 inches in the back, how much does the tip of a perpendicular rod on the plane move?

The image above shows the situation. There is a 2 foot long plane (width unknown) with a 3 foot high rod in its center perpendicular to the plane. and the plane in moved up 8 inches at the front, and lowered 8 inches from its original position. How much does the center of the circle move?

Also: how much would the center of the circle move if the front is only raised 8 inches, and the back is not lowered at all?

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When you move the plane up 8in at the front, where the pivot? The middle or (presumably) the back? –  Ian Coley Jun 3 '13 at 21:05
@FrankMcGovern The pivot will be in the middle –  Atomix Jun 3 '13 at 21:16

Such a device is impossible, or the picture is misleading. In the left picture, the blue distance is $\sqrt{12^2+8^2}\approx 14.4$ inches; after the "rotation" this same distance is filled by half of the 2 foot plane.

On the right picture, the blue distance had been filled by one foot of plane, but now there is an impossible right triangle with sides $8,12,12$ inches.

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For raising the front and lowering the rear, the angle of rotation is $\arcsin \frac 8{12}$ as you have a right triangle with hypotenuse (half the panel) $12$ inches and leg $8$ inches. The base of the post stays fixed and the post is tilted by that amount. The center of the ball then moves horizontally $24$ inches and down $\sqrt {36^2-24^2}\approx 26.8$ inches.
The approach is the same for the second part, but the pivot is in a different place. First figure out the location of the base of the post, then the location of the center of the ball.

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First problem: We assume that we are rotating the board about the base of the rod. Both supports will have to be moved inwards. A sketch shows that the angle of rotation is the angle $\theta$ whose sine is $\frac{8}{12}$.

Now we calculate the (straight line) distance from the old centre of the circle to the new centre. Imagine an isosceles triangle with sides $36$ inches, and angle between these sides equal to $\theta$. The distance moved by the centre is then the remaining side of the isosceles triangle. That is $(2)(36)\sin(\theta/2)$.

In our case, the calculator shows that $\theta$ is about $41.810315$ degrees. Take half of this, find the sine, multiply by $72$. We get about $25.69119$ inches.

Second problem: Here we assume that the point that remains fixed is the right end of the board, and we are rotating the board about that. Then we are rotating through an angle $\varphi$, where $\sin(\varphi)=\frac{8}{24}$. The left-hand support will have to be brought inwards,

Measure height from the plane of the pivot point. The old height of the centre was $36$. The new height is $4+36\cos(\varphi)$. This is about $37.941125$. So the height has changed by about $1.9411255$.

The centre is also displaced rightwards by an amount $36\sin(\varphi)=12$. By the Pythagorean Theorem, the distance between the old centre and the new is therefore about $12.155985$.

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