In fig 6 27 a box of cheerios left mass mc10 mathrmkgright...

In Fig. $6-27,$ a box of Cheerios $\left($ mass $m_{C}=1.0 \mathrm{kg}\right)$ and a box of Wheaties (mass $m_{W}=3.0$ $\mathrm{kg} )$ are accelerated across a horizontal surface by a horizontal force $\vec{F}$ applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.0 $\mathrm{N}$ , and the magnitude of the frictional force on the Wheaties box is 4.0 $\mathrm{N}$ . If the magnitude of $\vec{F}$ is $12 \mathrm{N},$ what is the magnitude of the force on the Wheaties box from the Cheerios box?

Transcript

00:01 Hi there, so for this problem we have two boxes as is shown in this figure.

00:05 The bus of wheatis has a mass of 3 kilograms and the other mass and the mass of chiro's has a mass of 1 kilogram and they both are accelerated across a horizontal surface and there is a force f that is applied to the shiro's both.

00:41 As is shown in the figure.

00:43 Now, the magnitude of the frictional force is also given for this problem, and for the chiro's bulbs, we have that that frictional force is equal to 2 meters, 2 newtons, and the frictional force for the witty spots is equal to 4 newtons.

01:05 Now, we also are given the magnitude of the force f that is being applied to the shiro's spots, that is equal to 12 newtons.

01:20 Now what we need to determine is the magnitude of the force on the witties spots for that shearer spot.

01:28 So we need to determine is that normal force of contact between the witty spots and the shiro spots.

01:42 Now in order to do this, what we need to do is to draw all of the forces that are acting on this and after that we can obtain for example first the acceleration and after of the system as a hold and after that apply in newton's second slot to isolating the mass of the vetoes so we can obtain that for of contact.

02:17 So that's precisely what we are going to do.

02:21 Now treating the two boxes as a single system we will have that a total mass for that as a total system is going to be something like this.

02:36 This as a hold with a force that is asserted to the right and a frictional force to the left.

02:51 The total mass of that is going to be equal to the sum of the masses.

03:00 So in this case we have 1 kilogram plus 4 kilograms.

03:05 So this will give us, oh sorry, 3 kilograms, 3 kilograms, and this will give us 4 kilograms for the total mass.

03:16 Now the total frictional force is the sum for h of them, which is the frictional force for c plus the frictional 4 for the widow's bobs.

03:32 So in here we will have the frictional force for c is 2 newtons and the other frictional force is 4.

03:42 So this will give us 6 newtons.

03:46 And that's it.

03:49 Now we can apply newton's second slow to attain this so we will have.

03:53 Since the force f is pointing to the right with 6.

03:56 That to be positive.

03:58 So some of the forces we will have the force f minus the frictional force and this is equal to the total mass times the acceleration.

04:10 So from this we can obtain the acceleration because that it will be the force minus the total frictional force over the total mass...

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