Question

A 493 kg mass is brought close to a second

mass of 171 kg on a frictional surface with
coefficient of friction 1.2.
At what distance will the second mass be-
gin to slide toward the first mass? Assume
that the two masses are made of a such a
dense material that their dimensions are very
small and one does not have to worry about
them touching before sliding begins. The
acceleration of gravity is 9.8 m/s² and the
value of the universal gravitational constant
is 6.67259 x 10-¹1 Nm²/kg².
Answer in units of mm.

Answer 1

The distance at which the second begins to slid toward the first mass, given that the coefficient of friction 1.2, is 0.053 mm

How to calculate the distance?

First, we shall obtain the static force existing. Here we shall use the second mass since it is resting on the surface. Details below:

• Mass (m) = 171 Kg
• Acceleration due to gravity (g) = 9.8 m/s²
• Normal reaction of first mass (N) = mg = 171 × 9.8 = 1675.8 N
• Coefficient of friction (μ) = 1.2
• Static force (F) =?

F = μN

= 1.2 × 1675.8

= 2010.96 N

Finally, we shall obtain the distance. Details below:

• First mass (M₁) = 493 Kg
• Second mass (M₂) = 171 Kg
• Gravitational constant (G) = 6.67259×10⁻¹¹ Nm²/Kg²
• Static force = 2010.96 N
• Force of gravity (F) = Static force = 2010.96 N
• Distance apart (r) = ?

`r = \sqrt{(GM_1M_2)/(F)} \\\\r = \sqrt{(6.67259*10^(-11)\ *\ 493\ *\ 171)/(2010.96)}\\\\r = 0.000053\ m\\\\Multiply\ by\ 1000\ to\ express\ in\ mm\\\\r = 0.000053\ *\ 1000\\\\r = 0.053\ mm`