A baggage conveyor is used to unload luggage from an airplane. The 10-kg duffel bag A is sitting

A baggage conveyor is used to unload luggage from an...

A baggage conveyor is used to unload luggage from an airplane. The $10-\mathrm{kg}$ duffel bag $A$ is sitting on top of the $20-\mathrm{kg}$ suitcase $B$. The conveyor is moving the bags down at a constant speed of $0.5 \mathrm{~m} / \mathrm{s}$ when the belt suddenly stops. Knowing that the coefficient of friction between the belt and $B$ is 0.3 and that bag $A$ does not slip on suitcase $B$, determine the smallest allowable coefficient of static friction between the bags.

Transcript

00:01 We have a conveyor, a baggage conveyor here, and a couple packages or luggage coming down it.

00:10 And it's coming down and all of a sudden this abruptly stops and we want to figure out, let's see here.

00:18 The upper bag, this one on top weighs 10 kilograms, this weighs 20 kilograms.

00:25 The coefficient of friction between the conveyor and the bag is 0 .3.

00:30 The incline is 20 degrees and it suddenly stops and knowing that the coefficient of friction, okay, the bag does not slip bag a, this does not slip on this.

00:42 The smallest allowable friction of static between the bags.

00:47 So basically you want to know what the friction forces, the coefficient of friction between these two bags, this one doesn't slip when this abruptly holds.

00:58 So i drew free body diagrams of each bag.

01:03 And we can see here that we have the weight, this is the upper bag, its weight, the normal from the lower bag, the friction force.

01:11 And then on the lower bag we have the normal from the upper bag, the friction force from the upper bag, it's weight, and then it's friction force and normal force between it and the conveyor.

01:26 Now we know that they're assuming that they're stuck together.

01:30 That's kind of the constraint on this.

01:32 So they have the same acceleration.

01:35 So we have free by diagram and we do balance of forces in this direction for this equals fb plus wb sine theta equals the mass of that package times its acceleration.

01:52 Summation of forces in this direction for the lower package gives us this.

02:00 Now let's see here the normal forces, the summation of forces in this direction here for this package gives us this equation and this equation for the lower package gives us this equation.

02:15 So what we can do is basically we just want to start eliminating stuff.

02:21 If we actually add these two equations, we simply get that this force drops out.

02:31 And basically this equation is the equation if we were to look at this as one system.

02:38 Because notice that fb as an internal force if we just look at this as a system.

02:44 And likewise, nb is an internal force...

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