Physics, 14.07.2021 21:20, anjoliej9

How does this work? Why won't it fall?
Why can't you continue building the tower forever?

From what I've been told, if the center of mass leaves it's support area, the object falls.
But the object on the tops center of mass is certainly outside of its support area (the object at the bottom)

Please explain! Thank you in advance. : )

How does this work?

Why won't it fall?Why can't you continue building the tower forever?From what

Answers: 3

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Answers

answered: naomijefferson22

Answer with explanation:

*Pre-condition: The mass of all blocks are evenly distributed.

As you add more and more blocks, the center of mass of the system changes.

Let the first block be the support. The maximum distance that can be unsupported (hung over) by the second block is equal to half the length of second block.

Why?

Let's take a look at our formula for torque. Torque, $\tau$ , is given by:

$\tau=rF\sin \theta$ , where is radius, is force, and $\theta$ is the angle between the radius and level arm.

$\sin \theta$ is used to calculate the relevant component of a force producing torque. In this case, the only force acting on the object is the force of gravity. Therefore, $\theta =90^{\circ}$ and recall $\sin 90^{\circ}=1$ .

Conceptually, it's more important to look at the term here. From our formula, we can see that if r=0 , there is no torque.

The point of pivot will be at the center of mass. Here's the important part:

As long as the point of pivot is supported, will remain zero and no torque will be created from the force of gravity. As you keep stacking blocks, as long as the center of mass of the entire system remains supported from your first support, the tower will not fall.

In the given picture shown, that first block will be your support.

The 6 blocks on top of that first block form a center of mass that is still on that first block, thus allowing the tower to remain standing.

answered: Guest

3x + 2x = both people using the same measure of like 5x = x= 25 meters per second. 25ms. it takes them 4 seconds to

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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How does this work? Why won't it fall?
Why can't you continue building the tower forever?