Box Mass and Friction: Calculating Time Before Fall

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SUMMARY

The discussion centers on calculating the time it takes for a 12.5 kg box to fall off a truck that accelerates at 2.19 m/s². The coefficients of friction are given as mu_s = 0.190 for static friction and mu_k = 0.150 for kinetic friction. It is established that if the truck's acceleration exceeds the maximum static friction force, only kinetic friction should be considered for the box's motion. This understanding simplifies the calculations significantly, allowing for a straightforward approach to determine the time before the box falls off.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to perform basic kinematic equations
  • Familiarity with force diagrams and free body diagrams
NEXT STEPS
  • Calculate the maximum static friction force using mu_s and the box's weight
  • Determine the net force acting on the box when the truck accelerates
  • Use kinematic equations to find the time until the box falls off the truck
  • Explore the effects of varying coefficients of friction on the box's motion
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in dynamics and frictional forces in motion scenarios.

trajan22
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A box of mass 12.5 kg rests on the flat floor of a truck. The coefficients of friction between the box and floor are mu_s = 0.190 and mu_k = 0.150 . The truck stops at a stop sign and then starts to move with an acceleration of 2.19 m/s^{2}.
If the box is a distance 1.79 m from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck?

my only question with this problem is how the static friction and kinetic frictional forces are related when calculating the final force...i know how to get each of these frictional forces just now how to put them together since kfriction acts over the entire movement while sfriction only acts for on instant
 
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Assume that truck begins its acceleration instantly. If the acceleration is enough to make the box slip along the floor of the truck, then you only need to consider kinetic friction when finding the forces on the box.
 
oh ok well that makes this problem infinitely easier than i thought
 

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