Force of Water on L-Shaped Tank Faces

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SUMMARY

The discussion focuses on calculating the forces exerted by water on the faces of an L-shaped tank. For face A, the force is calculated as FA = 2.5 x 106 N, leading to a net force of FAnet = 5 x 106 N when accounting for atmospheric pressure. For face B, the force is determined to be FB = 3.1 x 106 N, resulting in a net force of FBnet = 5.6 x 106 N. The discussion highlights the necessity of integration for accurate calculations, particularly for face B.

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The L-shaped tank shown in Figure 14-33 is filled with water and is open at the top.

hrw7_14-33.gif



(a) If d = 5.0 m, what is the force on face A due to the water?
N (up)
(b) What is the force on face B due to the water?
N (right)

The Force of water on the face A
FA = PA*AA
= rwg hA* AA
= rwg ( 2d ) * AA
0r FA = rwg ( 2d ) * d2
putting the values we get FA = 2.5*106 N
The atmospheric force on the face A = FA= ( 1.013*105 )*( 52)
= 2.5*106 N
Therefore Net force FAnet = Fatmp + FA
FAnet = 5*106 N

b) Force on the face B due to the water alone = FB
FB = PB*AB
= rwg ( 5d /2) d2
FB = 3.1*106 N
Thus, net force on face B FBnet = Fatmp + FBnet
FBnet = 5.6*106 N

I have given the problem a try. Please confirm the answer
 
Last edited:
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I'll leave this problem in Advanced Physics (Instead of Intro Physics) for now, because it appears to require an integration for (b).

Premed -- why do I think you need to use integration for part (b)?
 

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