Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg
Answer:
2.65 MPa
Explanation:
To find the normal stress (σ) in the wall of the basketball we need to use the following equation:
<u>Where:</u>
p: is the gage pressure = 108 kPa
r: is the inner radius of the ball
t: is the thickness = 3 mm
Hence, we need to find r, as follows:
<u>Where:</u>
d: is the outer diameter = 300 mm
Now, we can find the normal stress (σ) in the wall of the basketball:
Therefore, the normal stress is 2.65 MPa.
I hope it helps you!
Answer:
It is used for solving optimization problems in which,given some functional,one seeks the function minimizing or minimizing it.
Explanation:
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Syntax
Compiling the code, punctuation and characters must be corrected.
Answer:
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.
Explanation:
Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.
