Figure 6 32 shows three crates being pushed over a concrete...

Figure $6-32$ shows three crates being pushed over a concrete floor by a horizontal force $\vec{F}$ of magnitude 440 $\mathrm{N}$ . The masses of the crates are $m_{1}=30.0 \mathrm{kg}, m_{2}=10.0$ $\mathrm{kg},$ and $m_{3}=20.0 \mathrm{kg} .$ The coefficient of kinetic friction between the floor and each of the crates is 0.700 . (a) What is the magnitude $F_{32}$ of the force on crate 3 from crate 2$?$ (b) If the crates then slide onto a polished floor, where the coefficient of kinetic friction is less than $0.700,$ is magnitude $F_{32}$ more than, less than, or the same as it was when the coefficient was 0.700$?$

Transcript

0:00 Hi there.

00:01 So for this problem, we have three crates.

00:04 There are being pushed over a concrete floor with an horizontal force of magnitude 440 neutins.

00:13 So we are giving the information about the masses of each of these crates.

00:18 And we are also told about the coefficient of kinetic friction between the floor and each crate.

00:25 It is the same and it is 0 .7.

00:28 So the first question, in here is to calculate the force f3 -2 or the force on the crate 3 from the crate 2.

00:44 Do we need to calculate that force? so to do that, we are going to start by isolating the crate 3.

01:00 So i'm going to draw an access in here when we can draw all of these forces acting on this crate.

01:10 This is the y -adxes, this is the x -az.

01:14 So the forces opting on it are the normal force that we already call n3 because is the crate 3.

01:27 We also have a force f32, which is the force on this crate produced by the crate 2.

01:42 And we also have the weight three or the weight of this crate.

01:48 And we are, since the system is moving to the right, there is going to be a friction force that opposes that motion.

02:01 So we have a friction force to the left.

02:05 We're going to call this friction force three.

02:08 So when we sum the forces, well, if we start with the, forces in the wide direction, we're going to obtain that the normal force 3 is equal to the weight 3 because there are no other forces on that component.

02:24 And of course, this is equal to 0 because we don't have motion in that component, in that y component.

02:33 Now, for f of ads, for the forces on the component, adds, or the axis x, or the axis adds, we have the positive force 3 -2 minus the friction force 3.

02:52 And this is equal to the mass 3 times the acceleration.

02:59 Now the acceleration is the same for all of the crates because they are moving at the same time.

03:06 So we don't need to label that acceleration.

03:10 So we know that the definition of the friction.

03:13 Force is that it is equal to the normal force time the coefficient of kinetic friction, kinetic if the system is moving, like in this case.

03:23 So from this equation, we can obtain that the force 3 -2 minus the friction force, which in this case is the normal force three times the kinetic coefficient, and we obtain that the normal force is equal to the weight 3.

03:41 So we will have the weight 3 times the coefficient of kinetic friction.

03:48 And this is equal to the mass m3 times the acceleration.

03:54 So the force 3 -2 is going to be, we know that the definition of the weight is the mass of the grade times the acceleration due to gravity.

04:09 So we can solve for f -3 -3.

04:13 3, 2, and we know that this is going to be, we can factorize the mass m3.

04:21 We will have that this is g, musu k, and the coefficient, plus a.

04:36 So that is the expression for the force 3 -2.

04:41 As you can see, we don't know the value for the acceleration, so we need to find another way to find the acceleration to plotting in this and obtain the force f3 -2.

04:55 One way to obtain this is to assume that this system as a whole...

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