(II) A flatbed truck is carrying a heavy crate. The coefficien...

Question

(II) 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?

Accepted Answer

Hey, everyone in this problem, a heavy block is on the floor of a moving bus given that the coefficient of static friction between the floor of the bus and the block is 0.45. We are asked to calculate the maximum deceleration of the bus that won't cause the block to slide over. We're given four answer choices all in meters per second squared. Option A 0.046 option B 4.4 option C 9.8 and option D 44. So we are told about this coefficient of static where you're thinking about forces. And so what we're gonna do is draw a free body diagram. So we have our block, it is resting on the floor of a bus. So because it's resting on a surface, we know there's gonna be a normal force acting upwards. We also have the force of gray, it's gonna point to downwards. OK? And we're gonna assume that our bus is moving to the right, we have some speed to the right. And what that means is that we want the block not to slide forward. OK? When the bus decelerates. And so the force of friction that we have is going to act to the left. OK. It's going to oppose the motion and we are told that this is static friction. So we have FS pointing to the left. All right. So let's take up into the right to be our positive directions. We want to find the maximum deceleration given this friction coefficient that we have. So we want to think about a way to relate our forces to our acceleration. And we can do that through Newton's second law. So recall that Newton's second law tells us that the sum of the forces and in this case, we're talking about the horizontal or X direction is gonna be equal to the mass multiplied by the acceleration in that direction. So in this case, that's the acceleration of our bus. So the some of the forces in the X direction is equal to ma if we look at our diagram, OK. Our free body diagram, we can see we only have one force acting in that horizontal direction and that is the force of static friction FS and it's acting in the negative direction. So we have negative FS it's gonna be equal to the mass multiplied by the acceleration. Now, we wanna find the maximum deceleration and that's actually gonna correspond to our maximum static friction point. So we can write this equation as negative FS max is equal to the mass multiplied by the maximum acceleration or in this case, we're gonna look for the maximum deceleration. OK. So we're looking for a max here, this maximum static friction force recall, we can write that as the coefficient of static friction multiplied by the normal force. So we have negative us and is equal to M multiplied by a max. And we want to solve for a max. So we can divide both sides by M. We get that A max will be equal to negative in us. N divided by. Yeah. And we know the coefficient of static friction in the US. We don't know this normal force yet and we don't know this mass. And so what we're gonna do is we're gonna switch and we're gonna try to calculate this normal force and, and to do that, we need to look at the sum of the forces in the Y direction for our block because that's where that normal force is acting. And if we look in the Y direction and the sum of the forces is going to be equal to zero because the block is in equilibrium in that direction, it's not moving up and down at all. It's staying stationary in that direction, the sum of the forces will be equal to zero. The forces we have acting are the normal force in the positive direction, the force of gravity in the negative direction. So N minus FG is gonna be equal to zero. This tells us that the normal force will be equal to the force of gravity. And we recall that the force of gravity is just equal to the mass multiplied by the acceleration due to gravity. So we have N is equal to MG. So we have this expression for N. Now we can substitute it back into our equation for a max. So we have that a max will be equal to negative us multiplied by M multiplied by G all divided by M. And what we can see now is that this mass M is gonna divide it from our equation. So we don't actually need to know the mass. In this case, we just need that frictional coefficient and the acceleration due to gravity. OK? So let's substitute in those values and we can solve. OK. So our maximum acceleration will be negative 0.45 multiplied by 9.8 m per second squared. And this gives us an a max value of approximately negative 4.41 meters per second squared. OK? Now, that negative just indicates that we have a deceleration, which is what we were told in the problem. We're looking for the max deceleration. And so if we compare what we found to our answer choices, we're gonna take a positive value because when we call it a deceleration and we can drop that negative and we round to two significant digits, we find that the answer matches with option B 4.4 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

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(II) 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?

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