Part a consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average r
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Answer:
Explanation:
In the x direction the force will be
½(-w₀)L/2 = -¼w₀L
acting ⅔(L/2) = L/3 below the x axis.
In the y direction the force will be
½(-w₀)L + ½w₀L/2 = -¼w₀L
the magnitude of the resultant will be
F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛
in the direction
θ = arctan(-¼w₀L / -¼w₀L) = 225°
to find the distance, we balance moments
(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]
(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]
(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]
(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L
(√⅛)[d] = 5L/24
d = 5L/24 / (√⅛)
d = 5√⅛L/3
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
- The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.
Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, <em>C</em> represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
The steepness of the slope is therefore;
Where;
= Half the width of the truck =
= 1.2 m
= The elevation of the center of gravity above the ground = 2.2 m
The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.
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