Question

If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth will the timber stand vertically

Answer 1

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

                                   F_b = Y * V_timber

                                   F_b = 9810*0.15*0.15*x

                                   F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

                                   (M)_p = 0

                                   W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

                                   670*0.5*3.6 - x^2 * 0.5*226.7 = 0

                                   x^2 = 1206 / 113.35

                                   x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

                                   sin(Q) = y / x

                                   sin(Q) = 2.1 / 3.26

                                   Q = sin^-1 (0.6441718)

                                   Q = 40.1 degrees