Physics help?

The attraction force between 2 masses at a distance r apart is given as

F=-KMm/r^2, K=the universal constant. Now let

M=the mass of the mountain

m=the mass of the bob

r=12 km

The horizontal force of attraction of the bob by the mountain is found

F=-mg tan(0.5*)=-9.8m(0.00873)

-KMm/12000^2=-9.8(0.00873)m

=>

M=1.23*10^7/K=1.23*10^7/(6.67*10^-11)=1.844*10^17 kg approximately.

M = mass of mountain

m = mass of bob (he's gained weight)

F = force of gravity between bob and mountain

r = 12,000 m

G = 6.67×10^-11 Nm^2/kg^2

g = 9.8 m/s^2

F = GMm/r^2

tan(0.5°) = F / (mg)

M = gr^2 tan(0.5°) / G

I could ask how you know the plumb is pulled, as that is the local "vertical" ...  but anyway...

You may need the mass of the plumb bob, call it m, and the mountain M

Force of attraction or force of deflection F = mMG/d² = mM(6.674e-11) / (12000)²

tan 0.5º = force of deflection / force of gravity = 0.00873

0.00873 = [mM(6.674e-11) / (12000)² ] / mg

M(6.674e-11) / (12000)² = 0.00873(9.8)

M = 0.00873(9.8)(144000000) / 6.674e-11 = 1.85e17 kg

Gravitational attraction in newtons

F = G m₁m₂/r²

G = 6.674e-11 m³/kgs²

m₁ and m₂ are the masses of the two objects in kg

r is the distance in meters between their centers

All the values you need are right there