Two supports, made of the same material and initially of equal length, are 2.0 m apart. A stiff board with a length of 4.0 m and a mass of 10 kg is placed on the supports, with one support at the left end and the other at the midpoint. A block is placed on the board a distance of 0.50 m from left end. As a result the board is horizontal (that is, the downward force on each support is the same). The mass of the block is:

20 kg

Explanation:

Assuming that the board remains horizontal with the unknown mass on it, then F = kx

If we add the vertical forces to zero, we have something like this

2F - (M + 10)g = 0

2F = (M + 10)g, next, divide both sides by 2

F = (M + 10)g/2

Since we were able to sum the moments at the right end of the board to zero, we then proceed to find the unknown mass M

To start, we say, Let the clockwise moment is positive, and so

F * 4 + F * 2 - Mg * 3.5 - 10 * g * 2 = 0

4F + 2F - 3.5Mg - 10 * 2 * g = 0

6F - 3.5Mg - 10 * 2 * g = 0,

Remember from above, we say that

F = M + 10)g/2, now, all we do is substitute it inside this equation

6 * (M + 10)g/2 - 3.5Mg - 10 * 2 * g = 0

3 * (M + 10)g - 3.5Mg - 10 * g * 2 = 0, divide all sides by g(so as to eliminate it)

3 (M + 10) - 3.5M - 10 * 2= 0

3M + 30 - 3.5M - 20 = 0

-0.5M + 10 = 0

0.5M = 10

M = 10/0.5

M = 20 kg

the force required to move the cart is divided by three or 33.33 n.