m1= 443 kg, m2=181 kg
F(fr) = F =G•m1•m2/R²
μ•m2•g= G•m1•m2/R²
μ•g= G•m1/R²
R =sqrt(G•m1/ μ•g)
A 443 kg mass is brought close to a second
mass of 181 kg on a frictional surface with
coefficient of friction 0.5.
At what distance will the second mass begin to slide toward the first mass? The
acceleration of gravity is 9.8 m/s
2
and the
value of the universal gravitational constant
is 6.67259 × 10
−11
N · m2
/kg
2
.
Answer in units of mm
1 answer
1 answer