FIGURE P7.47 shows a 200 g hamster sitting on an 800 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μₛ = μₖ = 0 is sprayed on the top surface of the block, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

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Hey, everyone in this problem, we have a figure that shows a 250 g slab placed on a 950 g frictionless wedge. OK. The whole assembly rests on level ground. And we're asked to determine the normal reaction experienced by the wedge. If the slab is sliding downwards on the incline structure, we have four answer choices all in newtons. Option A 11.1 option B 9.31 option C 11.8 and option D 2.12. OK. So we have our diagram. It shows everything that we were told in the problem. Our slab sliding down this incline and the incline makes a 30 degree angle with the ground. OK. So that's gonna be important. So we're looking for the normal reaction. Yeah, the normal reaction. Well, that's gonna be the normal force experienced by the witch is we're looking for a force. So let's go ahead and draw our free body diagrams and, and we're gonna start with the slab. OK? So the slab we are gonna do in green because it's drawn in green in a diagram. OK. So first slab, what do we have we're gonna have a normal force acting because it's sitting on that wedge and it's gonna act perpendicular to the surface. So this is gonna be our normal force. And s for the lab, we have our force of gravity that's gonna have to directly downwards, straight downwards towards the ground. We're gonna call that FGS. And we can break this force of gravity up into two different components and we can break it up into the force of Grammy FGX. OK? For the slob and FGY for the slob. OK. And those are broken up into that tilted axis. All right. So that's for the lab. But what we're really interested in is this wedge. So let's go ahead and draw the free body diagram for the witch. Now, for the, we, it's gonna be very simple. We're gonna have a normal force pointing upwards now. OK? Because we're resting on this level ground. We have our normal force. NW We have the force of gravity acting straight downwards a FGW. What other forces do we have? Well, this slab is in contact with our witch. OK. So we're gonna have a force from that contact and that force is gonna be equal of magnitude in opposite to the normal force of the slab, that normal force of the slab is pointing upwards. And so this equal and opposite reaction is gonna be pointing downwards. OK? So this is gonna be the normal force due to the slap action reaction pair. OK. So let's write that in here. So that, that's clear we have an action reaction pair between these forces on the wedge and the slap. Now, for this moral force, we can break this up into its components just like we did. For the force of gravity on the slab, we're gonna break this up into the typical horizontal and vertical direction. So we have NSX and NSY. All right. And what we also know is that this mural force makes a 30 degree angle with that force of gravity with the vertical. All right. So we have our slab, we have our wedge, these free body diagrams drawn. Let's now move to look at the forces. OK? We want to calculate the normal force experience by the witch that's nw in order to do that, we're gonna have to look at the sum of the forces in the Y direction. And if we look at the sum of the forces in the Y direction, we have this normal force. We have the force of gravity which we can calculate because we know the mass, but we also have some portion. OK. The Y component of this normal force from the slab. OK. So we won't be able to solve for this normal force of the wedge right away. We have to first solve for the normal force of this love. So for the slab, we're gonna take the tilted axis. OK. So that tilted axis pointing upwards perpendicularly to the wedge is gonna be positive. And we're gonna look in the Y direction. OK? Because we want that normal force ns so the sum of the forces in the Y direction is going to be equal to zero. OK. We're told that the slab is sliding down the incline but it is not moving up and down relative to the incline. So in this tilted Y direction, it is not moving, it is in equilibrium, the sum of the forces is gonna be equal to zero. So the forces we have, we have that normal force and f acting in the positive direction. And we have this Y component of the force of gravity acting in the negative direction. So we have MS minus fgys is equal to zero. This tells us that the normal force and s is going to be equal to the force of gravity. The Y component of the slap, right, we're called the force of gravity is given by the mass. OK. So here the mass of the slab MS multiplied by the acceleration due to gravity. But here we're looking at just the Y component. So let's go back to our diagram here. Now, this lab makes an angle with the horizontal or I guess the wedge story makes an angle with the horizontal 30 degrees. And so the angle between the Y component of the force of gravity and the force of gravity itself is gonna be 30 degrees that Y component is the adjacent side to our angle. And so we're gonna be multiplying by cosine of 30 degrees. All right. Now, we can substitute in our values. We have the mass which is gonna be 0.25 kg. And we're told that it is 250 g. We want to use our standard unit to get to that unit of a Newton. So we divide by 1000 to go from grams to kilograms. So 0.25 kg multiplied by 9.8 m per second squared multiplied by cosine of 30 degrees. And this gives us a normal course. And s that is equal to 2.122 nos, all right. So we have this normal force. That's great. But remember this is not the normal force. We were looking for a, the questions asking us for the normal force experienced by the witch. What we needed to do was calculate the normal force um experienced by this lab first, because we have an action reaction pair with that force on the wedge. OK. So now we can get to the equation for the wedge. We have the sum of the forces, OK? And again, we're gonna look in the Y direction in the Y direction this wedge is not moving, OK, up and down. So the sum of the forces again is gonna be equal to zero. Now, what forces do we have acting in the wide direction on our witch. In the positive direction, we have the normal horse NW in the negative direction, we have the force of gravity. FGW. And we all know the Y component of the normal force. NS. All right. So putting that all together, we get NW minus FGW minus NSY is equal to zero. And this tells us that the normal force and W is going to be equal to fgwnsy. Now, a force of gravity just like before recall, that's gonna be the mass multiplied by the acceleration due to gravity. And in this case, it's the mass of the wedge. Now, the Y component of the normal force of the slab and going back to our diagram again, this makes an angle of 30 degrees with the vertical. OK. So when we break it into the Y component, we're looking at the adjacent side. So it's gonna be related through cos, all right. So we had and that normal force multiplied by cosine of 30 degrees. Substituting in all of our information, the mass of the wedge, just like the mass of the slab, we're gonna convert to kilograms by dividing by 1000 we get 0.95 kg multiplied by 9.8 m per second squared plus 2.122 newtons multiplied by cosine of 30 degrees. And when we work this all out, we get that the normal force experienced by the wedge is gonna be 11.148 newtons approximately. We're gonna go to our answer choices and we are gonna round this to three significant digits like the answer choices and we can see that the correct answer is gonna be option a 11.1 Newtons. Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Friction

Newton's Second Law

Weight and Normal Force