Torque on a Dam

Classical Mechanics Level 3

Water exerts a force on the dam. It also exerts a torque about the axis passing through O in the figure. The net force the water exerts on the dam can be modeled to act at a height a b H \frac{a}{b}H above O (So the torque exerted by the water is a b F H \frac{a}{b}FH ). What is a + b a+b ?

You can neglect the atmospheric pressure as it acts on both sides of the dam

This Problem is taken from Physics for Scientists and Engineers with Modern Physics


The answer is 4.

Saad Haider by Saad Haider , 22, Singapore

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1 solution

Saad Haider
Mar 16, 2014

Force acting on the area w d y wdy is

d F = P d A = P w d y = ρ g ( H − y ) w d y dF = P dA = Pw dy = \rho g(H-y)w dy

∫ d F = ∫ 0 H ρ g ( H − y ) w d y \int dF = \int_{0}^{H} \rho g(H-y)w dy

F = 1 2 ρ g w H 2 F = \frac{1}{2}\rho gwH^{2}

Next, we calculate the net torque about O.

d τ = y d F = P y d A = ρ g ( H − y ) w y d y d\tau = y dF = Py dA = \rho g (H-y)wy dy

∫ d τ = ∫ 0 H ρ g ( H − y ) w y d y \int d\tau = \int_{0}^{H} \rho g (H-y)wy dy

τ = 1 6 ρ g w H 3 \tau = \frac{1}{6}\rho gwH^{3}

Now we model the net force we calculated earlier on to act at a distance h above O to produce this torque

F h = τ Fh = \tau

1 2 ρ g w H 2 h = 1 6 ρ g w H 3 \frac{1}{2}\rho gwH^{2}h = \frac{1}{6}\rho gwH^{3}

h = 1 3 H ⇒ 1 + 3 = 4 h = \frac{1}{3}H \Rightarrow 1 + 3 = 4

4 Helpful 0 Interesting 0 Brilliant 0 Confused

I do it in this way.

Arghyanil Dey - 7 years, 1 month ago

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