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RevilewvPar _4200 kg tuck paiked onslope .How big the friction force on the truck? The coefficien of static friction betrcen the tires and the road Express your 4ng...

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RevilewvPar _4200 kg tuck paiked onslope .How big the friction force on the truck? The coefficien of static friction betrcen the tires and the road Express your 4nghei [Ko significant figures and irclude the appropriate units_You may want revey (Pages 141 145) _ValueUnitsSubrilRequest AnswerProvide Feedback

Revilewv Par _ 4200 kg tuck paiked on slope . How big the friction force on the truck? The coefficien of static friction betrcen the tires and the road Express your 4nghei [Ko significant figures and irclude the appropriate units_ You may want revey (Pages 141 145) _ Value Units Subril Request Answer Provide Feedback



Answers

$\mathrm{A} 4000 \mathrm{kg}$ truck is parked on a $7.0^{\circ}$ slope. How big is the friction force on the truck?

First will draw a free body diagram here. I suppose this is a ramp which is making in both Kita here in the heart gentle. And this is a truck. I suppose this is struck. Uh huh. Here am devil will be acting on this in the downward direction. And mm, this confident off MGI and these antifa this company and will be mg cost hater. In this question, we have to find how big imperfection force on the truck since this truck is part on a ram. So here mg scientific accompaniment off its weight. Try toe, pull this truck in this direction. But the fiction force acting in opposing direction and the magnitude off this friction force should be equal to the, um this anti ETA. So you can right here. The friction force should be equal toe mg sign Peter here. The mass of the crackers 4000 kg and the G due to gravity is 9.81 and the three days seven degree. After calculating this, you'll get the value of forces 47 82 15 Newton

So in this question, we've got a 4000 kilogram truck. It is parked on a slope of 15 degrees and were asked to calculate the fiction force on the truck. So presumably, if the truck is parked, we know that the velocity is going to be zero, and therefore the acceleration is going to be equal to zero. So that's the first thing that I would just make me know down Ah ah, when you're going to solve this problem. And the second thing I would do is draw picture. So what I've got here is the 15 degree slope, and I've labeled my coordinate system here. So why is going to be perpendicular to the slope and acts will be the direction that is parallel to the slope. It's just a little bit more convenient to use that particular coordinate system rather than just vertical horizontal s. So what I'm gonna do is I'm gonna draw my truck and I'm gonna label the forces on it. So I'm doing a little free body diagram here. The normal force will be perpendicular to the incline. The gravitational force, of course, will always point straight down towards the ground and then we will have a friction force that is a balancing the downward pulling force from gravity. So gravity is trying to pull the truck down the slope and the friction forces there to balance it out. Now, one thing that's gonna be important here is that the gravitational force does have some components in this particular, um, coordinate system. So if the coordinate system is kind of turned like this, rotated like this than the gravitational force will have an F G, ex components and an F G. Why component now? Ah, One thing to know is, if your incline is 15 degrees than this angle here, right at the top of this triangle will also be 15 degrees. So that's something that you can memorize. Ah, in order to help you to resolve components on inclined planes, if you would like. Now, in order to find with that friction forces, I need to take a look at what's going on in the X direction here. So let's look at the force bounce equation in the X direction, so F Net X is going to be equal to M A X. And as already noted, the acceleration for the truck will be zero. And let's take a look at the forces that are acting in the extraction. So of course, we have the X component of gravity and then opposing that in the negative direction will be that friction force. So here we go. We've created a equation for the friction for us. We just need to know what the X component is of gravity. So gravity is mg and we need to multiply by other sign 15 or Khost 15 to get the, um, component of gravity. Weaken C F G X is opposite from the 15 degrees in this triangle, so we need to use sign. So we plug and the trucks mass and the acceleration due to gravity multiplied by sine 15 degrees. And that will give us a friction force off about 10,000 Newtons. I'm just gonna keep one significant figure, cause there was only one significant figure in the mass. And this is the friction force. So typically speaking, um, the friction force, the static friction force will increase until it can increase anymore. So presumably, if the truck is stationary, then we're below that blower limit for the static friction force, the the maximum, um, static friction force

This question covers three concepts. First concept says that the maximum static friction force is less than the product of the coefficient of static static friction and the normalization was, and the acceleration provided by the static friction uh mostly less than or equal to the friction force upon to mass. And the third concept is of work Nigeria. Um It's strange that the network done is equivalent to the change. It cannot take much. So first let's draw the fretboard diagram for the stack and let's say the mass of the stack is m. So the forces acting on the status uh Wait, which is acting downward. Yeah, a normal reaction force which is acting upward, and a friction force acting along the direction of motion which provides the uh with bright the acceleration. Okay, so to solve discussion first, we need to convert the final speed of the truck into me just once again. So finally, speed is 56.6 mph. Or we can write the fund speeders 56.6 into 1609 Upon 3600 meters for a second. That the final speed of the practice ah 25 13 m/s. And now let's go with the acceleration of the trip. The acceleration of the truck is the final final speed. My first initial speed and the final speed is 25.3 m pose. Again, an initial speed is zero upon the accelerating time. That is 22.9 seconds. So from this, the acceleration After truck is 1.1 m or 2nd square so and stop acceleration. But what about the static friction is greater than the maximum value is greater than the acceleration of the trip. Then the stack up cement will not slip. Now there's Calcutta, let's calculate the yeah, maximum acceleration due to the static friction since the strike is not moving along the vertical direction. So we can say the normal reaction force is a cuBA into the weight and the maximum static friction is that you were into the coefficient of friction us into the normal reaction force that is empty. And from this we can say the maximum acceleration provided by the static friction. His newest times cheap. Let's substitute the value to calculate the maximum static friction. Uh Sorry, maximum static acceleration And the coefficient of friction is zero 372 for static friction In 29.81 m 4 2nd square. And we find the maximum acceleration that can be imported where the static friction is 3.65 m four seconds square, which is greater than the acceleration of the truck. So from this argument we can say that the cement stack will not slip. Okay because the ah friction is less than the maximum static friction. No, from the work energy firm. The work done by the gravity plus the work done by the normal reaction force plaster work done by the friction force. Is it your parents? The change in kinetic energy. Okay. And since the gravity is acting perpendicular to the displacement, therefore the work done by the gravity of zero, similarly, the work done by the normal reaction forces zero plus the work done by the friction force. And that is a Cuba. And to uh W. F. And which is the government of the change in kinetic energy, that is half of the mass of the state, that is 1143 0.5 kg into the square of the final spirit. That is 25.3 meters per second squared Man extermination contractors here zero. So the work done by the friction is secure into 3.66 in two tenders, 5 jules. Or we can write this as The work done by the friction force on the stack is 366 kg jewels.

This question covers three concepts. First concept says that the maximum static friction force is less than the product of the coefficient of static static friction and the normalization was, and the acceleration provided by the static friction uh mostly less than or equal to the friction force upon to mass. And the third concept is of work Nigeria. Um It's strange that the network done is equivalent to the change. It cannot take much. So first let's draw the fretboard diagram for the stack and let's say the mass of the stack is m. So the forces acting on the status uh Wait, which is acting downward. Yeah, a normal reaction force which is acting upward, and a friction force acting along the direction of motion which provides the uh with bright the acceleration. Okay, so to solve discussion first, we need to convert the final speed of the truck into me just once again. So finally, speed is 56.6 mph. Or we can write the fund speeders 56.6 into 1609 Upon 3600 meters for a second. That the final speed of the practice ah 25 13 m/s. And now let's go with the acceleration of the trip. The acceleration of the truck is the final final speed. My first initial speed and the final speed is 25.3 m pose. Again, an initial speed is zero upon the accelerating time. That is 22.9 seconds. So from this, the acceleration After truck is 1.1 m or 2nd square so and stop acceleration. But what about the static friction is greater than the maximum value is greater than the acceleration of the trip. Then the stack up cement will not slip. Now there's Calcutta, let's calculate the yeah, maximum acceleration due to the static friction since the strike is not moving along the vertical direction. So we can say the normal reaction force is a cuBA into the weight and the maximum static friction is that you were into the coefficient of friction us into the normal reaction force that is empty. And from this we can say the maximum acceleration provided by the static friction. His newest times cheap. Let's substitute the value to calculate the maximum static friction. Uh Sorry, maximum static acceleration And the coefficient of friction is zero 372 for static friction In 29.81 m 4 2nd square. And we find the maximum acceleration that can be imported where the static friction is 3.65 m four seconds square, which is greater than the acceleration of the truck. So from this argument we can say that the cement stack will not slip. Okay because the ah friction is less than the maximum static friction. No, from the work energy firm. The work done by the gravity plus the work done by the normal reaction force plaster work done by the friction force. Is it your parents? The change in kinetic energy. Okay. And since the gravity is acting perpendicular to the displacement, therefore the work done by the gravity of zero, similarly, the work done by the normal reaction forces zero plus the work done by the friction force. And that is a Cuba. And to uh W. F. And which is the government of the change in kinetic energy, that is half of the mass of the state, that is 1143 0.5 kg into the square of the final spirit. That is 25.3 meters per second squared Man extermination contractors here zero. So the work done by the friction is secure into 3.66 in two tenders, 5 jules. Or we can write this as The work done by the friction force on the stack is 366 kg jewels.


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