Question

Two boxes are sliding along a floor after being given an initial speed \[v\] to the right. the bottom box has a mass that is \[3\] times larger than the top box. the coefficient of kinetic friction between the bottom box and the floor is \[\mu_\text{floor}\]. the maximum coefficient of static friction between the boxes, \[\mu_\text{box}\], is \[0.13\].

Answer 1

The maximum value of μfloor that will allow the boxes to stop without slipping is 0.52

Analyze the forces:

Friction between bottom box and floor: μfloor * m_total * g (to the left)

Friction between boxes: μbox * m_top * g (to the left, acting on top box only)

Set up equations for motion:

Top box: F_net = m_top * a (friction between boxes is the only horizontal force)

Bottom box: F_net = (m_total) * a (friction between floor and both boxes is the net force)

Consider maximum static friction:

To prevent slipping, the friction between boxes must not exceed μbox * m_top * g.

Solve for μfloor:

From the top box equation: μbox * m_top * g = m_top * a

a = μbox * g

Substitute a into the bottom box equation: μfloor * m_total * g = (m_total) * μbox * g

μfloor = μbox * (m_total / m_total) = μbox * (1 + 3) = 4 * μbox

μfloor = 4 * 0.13 = 0.5