Find coefficient of static friction of a uniform box?

Question

A uniform box of mass 4.0 kg that is five times as tall as it is wide rests on the floor of a truck. What is the maximum coefficient of static friction between the box and floor so that the box will slide toward the rear of the truck rather than tip when the truck accelerates with acceleration a = 0.60g on a level road?

µs < _____

Answer 1

so the trick to this question is to balance the the torques on the box caused by gravity and the forward acceleration of the truck.  

Lets say the box is 5m tall and 1m wide.

The center of mass of the box will obviously just be the center of the box or 2.5m up and .5 meters from the side.

when the box first starts tipping the torque from gravity in the y direction will be.

Ty = F * D = m * a * D = (4kg)(9.8m/s/s)(0.5m) = 19.6 Newton-meters

now if we don&#039;t want the box to tip, the torque in the x direction must equal that of the y so we are going to find the acceleration necessary to cause this torque:

Tx = F * D = m * a * D = (4kg)(a)(2.5m) = 19.6 Newton-meters

Solving for a we get:

a = 1.96m/s/s

This is the maximum acceleration the box can undergo without tipping over.

The truck is accelerating at 0.6 * g  = (0.6)(9.8m/s/s) = 5.88m/s/s

since we only want the box to experience 1.96m/s/s of this acceleration we can take the coefficient of friction to be the ratio between the two:

µs &lt; (1.96m/s/s) / (5.88m/s/s) = 0.333 = 1/3

µs &lt; 1/3