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(II) Two crates, of mass 65 $\mathrm{kg}$ and 125 $\mathrm{kg}$ , are in contact and atrest on a horizontal surface (Fig, $32 ) . \mathrm{A} 650$ -N force is exertedon the 65 -kg crate. If the coefficient of kinetic friction is $0.18,$calculate $(a)$ the acceleration of the system, and $(b)$ the forcethat each crate exerts on the other. (c) Repeat with the cratesreversed.

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a. $1.65^{m} / 5^{2}=a$b. $F_{I}=430.5 \mathrm{~N}$c. $=221.9$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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in this problem, you have two crates of mass 125 and 65 kilograms, with a force of 650 Newtons being applied to the 60 65 kilogram mass and a coefficient of friction of 0.18 between the two blocks and the ground and were asked to calculate the acceleration of the system. For part came and the force between the crates and party from part came going to start by drawing everybody diagrams for both of being objects and sort of the 65 kilogram mass. We have a normal force force of gravity. We have the applied force and then we have the force between two messes. So for this and the friction and then so true, to treat it as a system, we're gonna rent the sum of the forces in the ex direction. And so if we do that, these two forces cancel out and we're left with the applied force minus the friction on one block minus the friction of the other block, and then be two masses combined. So we know friction is our normal force times air coefficient of friction and our normal force is our mass times gravity for each object. So we have our fight force. My coefficient Times 1 25 times Gravity minus my coefficient times 65 times gravity all divided by my two masses, which is 1 25 plus 65 are by force was 650 mutants and that gives us an acceleration of 1.65 meters per seconds. Word for part B, or ask to solve for the force in between the two masses. So for this we need to go back to looking at just one of the three body diagrams. I think this one would be the easiest you could do it with. Either way, and if we look at the sum of the forces we have the force between the two masses might miss. The friction on just one of the masses equals the mass times acceleration. So we're just gonna be using the mass of the 125 kilogram in this office again. This problem can be redone using the other mess. But I think this was just easier because it's one less force than together Then this is 1 25 the acceleration that was sold for earlier Our coefficient of friction times o r mass times gravity. And this gives us a force between the two objects of 430.5 minutes. Part see asks us to repeat with the crates reversed. So instead of playing the 650 Newtons to the 65 kilogram mess, we're gonna buy it to the 125 kilogram Mets. Now, if you look at it, um, that means our this applied force will now be going to the left. But then this object will be moving to the left, and then my two frictions would just change direction, so this would still be true. So we still have the same acceleration that we did before. Now, instead, we're now going to have different forces acting between the two masses. So I'm gonna again use that 125 kilogram mass just for to show what's going on again. We have that 6 50 that's now being applied. We have my friction going to the right, and then we have the applied force between the two masses. So this is what we're trying to solve for, so I'm gonna change my direction. So I'm gonna say going to the left is now positive. This makes our math a little easier and putting in our numbers again. My applied force was 650. My friction was my coefficient times my normal force. My mass again were amusing. Be 125 kilogram mass tends My acceleration from before gives me a new force in between the two masses of 221 point nine news.

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