2. A person is pushing box A across the floor at constant...

2. A person is pushing box A across the floor at constant speed. Atop box A is box B and next to and touching box A is box C. Box B Box A Box C Each box has a different mass with MA > MB > MC and there is friction between all surfaces with ?s > ?k. As the person pushes box A, all three boxes move together across the floor at constant speed. a) Sketch three free body diagrams - one for each individual box. Each force arrow should be labeled with the type of force (kinetic friction, static friction, normal, etc.) and the object exerting that force. [This is the same way we've been labeling forces all unit.] b) Rank all of the horizontal forces in order of increasing magnitude. Some forces may have the same magnitude so you answer will look something like F1 > F2 = F4 > F3 except that instead of 1, 2, 3, and 4 the labels will be the type and object labels from your diagrams in part a and you may have more or less than four horizontal forces.

Transcript

00:01 In this given question, there are three boxes, box a, b and c, that it represent by its masses with m a, mb, and mc.

00:16 Now one can see here that the box b is at the top.

00:21 Box a is below that and box c is adjacent to box a.

00:27 Now it has been given that a person is pushing box a across the floor at constant space.

00:33 That means the force exerted by the force exerted by the person on the boxes are equal to the sum of the friction forces objecting or let's say resisting the motion.

01:00 So one can write force equals to let's say f1 plus f2, sorry.

01:09 F1 plus f2 where f1 is the friction due to the mass of the boxes a and b and f2 is the friction frictional force between box c and the ground so let's first mention the forces in the fbd diagram or the free body diagram so box b bd or the free body diagram so box b due to its weight, a downward force will act that will be mbg and a counter normal reaction will act on the box b.

01:57 Let's say it is nb.

02:01 Let's represent nb here.

02:07 Now, one can also see here that a frictional force will be acting between these two surrogens.

02:17 Between the surface of the lower surface of the box b and the upper surface of the box a.

02:29 And if we look carefully, the main driving force for the box b in horizontal direction is basically the frictional force.

02:41 So it will be in this direction and let's represent it by fb, where fb is equal to mu s into normal reaction nb now if we equate here the forces in vertical direction that means nb minus m bg equals to 0 because in vertical direction there is no motion so here nb will be equals to m bg so from here one can conclude that frictional force fb will be mu s into to nb where we can replace nb with m bg now in horizontal direction there will also be a inertial force acting on the box b and that will be opposite to this and that will be equals to m b into a where a is the acceleration of box b in horizontal direction so one can write f b minus m b equals to mb a dash where a dash is the net acceleration of box b or let's say the net acceleration of the system where the system here represents is all box a b and c now here it has been given that the all this system is moving in horizontal direction with constant speed, that means a -dash will be equals to 0.

04:41 So here it will be 0.

04:47 And if we further solve it, we will get fb equals to mba.

04:55 That means the opposite force acting on the box b due to its inertia, that is equals to this frictional force acting in the right -word direction.

05:10 So these two forces are same in magnitude and opposite in direction.

05:19 So this is a free body diagram for the box b.

05:23 Now let's come to the free body diagram for box a.

05:34 Now here let's draw the box b here.

05:42 Let's imagine that these are the lower particle.

05:46 Or the part of the lower part of the box b.

05:51 Now due to inter -atomic interaction, the particles of box a was basically pulling this box b.

06:01 That's why the frictional force on box b was acting in right -word direction.

06:07 But in this case, due to the newton's third law, at the same time, the lower surface or the lower surface or the the inter -atomic particles of the box b was actually pulling the upper surface of box a in leftward direction where the friction is equal to f sorry the force is equal to in magnitude and that is equal to fb so for box b it was fb where fb equals to mu s into m b so here also fb will the force fb acting on the box a will be also equals to fb equals to mu s into mb into g and here mu s is written rather than mu k because there is no any kind of relative motion between box b and box a...

Question

Please give Ace some feedback