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
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Answer:
38.3 m/s
Explanation:
To find vertical component of initial velocity, you'd have to use sine ratio:
is vertical component of initial velocity and 
is initial velocity given which is 50 m/s.
A stone is projected at an angle of 50 degrees so 
= 50°. Substitute in the formula:
Therefore, the vertical component of initial velocity is approximately 38.3 m/s
(The picture is also attached for visual reference!)
Explanation:
The angular momentum is doubled because it is proportional to the angular velocity.
Fk = μK N
N = m a
N = 4 × a
N = 4a
Fk = μK N
17 = 5 × 4a
17 = 20 a
a = ¹⁷/₂₀ = 0.85 m/s²
