Answer :
Answer:
Explanation:
given,
length of lever = 1 m
diameter of cylinder = 25 cm
weight of cylinder = 200 N
hydrostatic force
[tex]F_H=\gamma hA[/tex]
= [tex]9810\times 10\times \dfrac{\pi}{4}1^2[/tex]
= 77048 N
now,
[tex]y_{cp}-\bar{y}= \dfrac{I}{\bar{y}A}[/tex]
= [tex]\dfrac{\dfrac{\pi r^4}{4}}{10\times\dfrac{\pi D^4}{4}}[/tex]
= 0.00625 m
Finding the resultant force
[tex]F = F_{buoyancy} - W[/tex]
[tex]F = \gamma_{water}A(10-l)-W[/tex]
[tex]F = 9810\times 10\times \dfrac{\pi}{4}0.25^2\times (10-l)-200[/tex]
F = 4615.5-481.5 l
taking moment about hinge
[tex]F_{H}\times 0.00625 - 1 \times F = 0[/tex]
[tex]77048\times 0.00625 - 1 \times(4615.5-481.5 l) = 0[/tex]
l = 8.58 m
The length of the chain such that the gate is just on the verge of opening is mathematically given as
l=8.58m
What is the length of the chain?
Generally, the equation for the is mathematically given as
[tex]F'=\gamma hA[/tex]
Therefore
[tex]9810*10* \dfrac{\pi}{4}1^2[/tex]
Fh= 77048 N
Where
[tex]ycp-y=\frac{\frac{\pi r^4}{4}}{10*\frac{\pi D^4}{4}}[/tex]
ycp-y=0.00625
In conclusion, resultant force
x = F'' - W
x = 9810* 10*( \pi/4 )*0.25^2 *(10-l)-200
x = 4615.5-481.5 l
Therefore
77048* 0.00625 - 1 *(4615.5-481.5 l) = 0
l=8.58m
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