Alightweight diving board is supported by two pillars. one is at the end of the board and the other is 3.30 m away. a 75-kg person is standing on the board at such a location that the pillar at the end (pillar #1) of the board exerts a 810-n force downward. determine the following. (a) distance the person is standing from the end pillar (pillar #1)

(b) magnitude of the upward force exerted by the second pillar (pillar #2)

a) x=3.63m  b)

Explanation:

In order to solve this problem we can start by building a free body diagram of the board and the pillars. (See attached picture).

a)

So in order to find the distance between the second pilar and the person x, we can do a summation of moments about the second pillar so we get the following:

which will look like this:

we can now solve for x (the distance between the second pillar and the person) so we get:

We know that the weight of the person is given by mg so the equation can be rewritten like this:

When substituting the respective data we get:

which yields:

x=3.63m

b)

in order to find the magnitude of the upward force exerted by the second pillar we can do a sum of forces in the vertical direction so we get:

which yields:

which can be solved for :

When subsstituting the data we get:

so the force exerted by the second pillar is 1.55kN.

What do you need on

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.