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## Homework Statement

The gravitational attraction due to a nearby mountain range might be expected to cause a plumb

bob to hang at an angle slightly different from the vertical. If a mountain range could be

represented by an infinite half-cylinder of radius a and density [itex]ρ[/itex] lying on a flat plane (see

attached image), show that a plumb bob at a distance [itex]r_0[/itex] from the cylinder axis would be deflected

by an angle [itex] θ ≈ \frac{πa^2Gρ}{r_0g}[/itex]. You can assume that [itex]θ << 1[/itex] and [itex]r_0 >> a[/itex]. In actual measurements

of this effect, the observed deflection is much smaller. Next assume that the mountain range

can be represented by a cylinder of radius a and density ρ which is floating in a fluid of density

[itex]2ρ[/itex] as illustrated. Show that the plumb-bob deflection due to the mountain range is zero in this

model. Since the latter result is in much better agreement with observations, it is postulated that

mountains, and also continents, are in isostatic equilibrium with the underlying mantle rock.

## Homework Equations

[itex] ψ_g=\iint \vec{g_c}\cdot\vec{dA} = -4πρG [/itex] and [itex] ∑τ=0[/itex]

## The Attempt at a Solution

**Part A)**First I need to calculate the gravitational field [itex] \vec{g_c} [/itex]. Because the mountain is approximated as an infinite half cylinder the only contribution comes from the thin slice of directly in front of the plum-bob.

[itex] ψ_g = \iint\|\vec{g_c}\|\|\vec{dA}||cosθ = -4πρG →\|\vec{g_c}\| \iint \|\vec{dA}\| = -4πρG → \|\vec{g_c}\|π(r_0) = -4πρG → \|\vec{g_c}\| = \frac{-4ρG}{r_0} \Rightarrow \|\vec{F_c}\| = \frac{-4ρGm(πa^2)}{2r_0}[/itex]

Now the other force acting on the bob is the Earth's gravitaional force given near surface of the earth is: [itex]\|\vec{F_g}\|=-mg[/itex]

Now consider the torque given by both forces acting on the plum-bob: [itex]∑τ=0 → -Lmgsin(θ)-L[\frac{-2ρGπa^2}{r_0}]sin(Θ) = 0 [/itex] where [itex]θ+Θ=90°[/itex] so [itex]θ<<1 \Rightarrow Θ≈90 [/itex] then the above equation simplifies to: [itex] mgθ≈\frac{2ρGπa^2m}{r_0} → θ≈\frac{2πa^2Gρ}{r_0 g}[/itex] and since the angle is small to begin with the factor of 2 can be neglected so I get the predicted result

**Part B)**I'm not really sure where to go with the next part so any hints would be greatly appreciated. What I do know: The earth's gravitation field is much stronger than the field due to the infinite half cylinder's (mountain's) and there is a bouyant force acting on the plum bob opposite to the earth's gravitational field. This lead me to believe there was a balancing viscous force due to the fluid, but I'm unsure of this.

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