A flatbed truck is carrying a heavy crate. The coefficient of static friction between the crate and the bed of the truck is 0.75. What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab of the truck?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. With this truck originally going to the right in the positive x direction say, and then decelerating, that means there's gonna have to be a force pulling back on the crate which is in the flatbed and that's going to be the friction force, it does that. And the friction force is going to equal mass times acceleration because it's the net force in the x direction. And we also know that the friction force is coefficient of static friction times the normal force and the normal force is going to equal gravity since there's, you know, the total up forces have to equal the total down forces in order for there to be no vertical acceleration in the crate. And so, we can substitute mg in place of the normal force. So, the friction force is muS mg, and the friction force is also negative ma because the friction force is to the left, and so I'd like to put negative signs explicitly in front of my force terms, and then, well, multiply both sides by negative one, and you get friction force is negative times mass times acceleration. And then you equate this with this and you get negative ma is muS mg. And divide both sides by m and the m's cancel, giving us that acceleration is negative muS times g. So, it's negative 0.75 times 9.8 meters per second squared which is negative 7.4 meters per second squared is the maximum acceleration. Or deceleration I guess you could say, that the flatbed truck could have to avoid having the crate slide.