Question

Chapter 13, Problem 018 Flying Circus of Physics Mountain pull. A large mountain can slightly affect the direction of "down" as determined by a plumb line. Assume that we can model a mountain as a sphere of radius R 1.00 km and density (mass per unit volume) 2.6 × 103 kg/m3. Assume also that we hang a 0.750m plumb line at a distance of 3R from the sphere's center and such that the sphere pulls horizontally on the lower end. How far would the lower end move toward the sphere? Number Units the tolerance is +/-2% SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM MATH HELP Question Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

T cos θ Fg 30 Tsin θ

Density of mountain $\rho=2.6*10^3\,kg/m^3$

Radius of mountain $R=1.00\,km=10^3\,m$

Mass of mountain $M=\rho V=\rho*\frac{4}{3}\pi R^3$

The sphere (mountain) pulls the plumb line,

Forces acting the plumb mass $m$ are,

i) gravitational force $mg$ , vertically downwards

ii) Components of tension $T\cos\theta$ vertically upwards, $T\sin\theta$ horizontally towards right.

iii) Force of attraction due to mountain (sphere) $F_g=\frac{GMm}{(3R)^2}$ horizontally towards left.

Net force acting on the plumb bob is zero, $F_g=T\sin\theta$ and $mg=T\cos\theta$

$\frac{T\sin\theta}{T\cos\theta}=\frac{F_g}{mg}$

$\tan\theta=\frac{GMm}{mg(3R)^2}$

$\tan\theta=\frac{GM}{9gR^2}$

$\tan\theta=\frac{G\rho *\frac{4}{3}\pi R^3}{9gR^2}=\frac{4\pi G\rho R}{27g}$

$\tan\theta=\frac{4\pi *6.67*10^{-11}*2.6*10^3*10^3}{27*9.81}=8.23*10^{-6}$

From the figure, $\tan\theta=\frac{x}{l}$ ( where $l$ is length of plumb line, $x$ is the distance moved by lower end of plumb line towards sphere )

$\Rightarrow x=l\tan\theta=0.750*8.23*10^{-6}=6.17*10^{-6}\,m$