Boxes are moved on a conveyor belt from where they are filled to the packing station 11.0 m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accelerates for half the distance, then decelerates for the final half of the trip. If the coefficient of static friction between a box and the belt is 0.56 what is the minimum transit time for each box?

Question

matt101 · Answer

Picture this as an equilibrium question. We want the net horizontal force to be zero so that way the box doesn't slide away. Static friction is the obvious horizontal force, but what is the force it's balancing? The force applied by the conveyor belt on the box, which is ma. So now we can set up an equation:

$$f=ma$$$$\mu mg = ma$$$$\mu g = a$$$$a=(0.56)(9.8)=5.488$$
Notice that the mass drops right out of the equation, leaving only the acceleration - exactly what we want! In this case, the maximum acceleration (and deceleration) of the conveyor built is 5.488 m/s^2. Consider just the acceleration - half the distance is 5.5 m. We can now calculate the time this half of the trip takes:

$$d= v_i t+ \frac{1}{2}at^2$$$$5.5= \frac{1}{2}(5.488)t^2$$$$t=1.42$$
The time it takes to cover half the distance is 1.42 s (we can disregard the negative value of t because it makes no sense here). That means the time to cover the full distance is twice that, 2.84 s!

Let me know if that makes sense!