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Thesolid; uniformly dense blocks of wood shawn below are both slovtly tilted differcnt helghtsincressing angle untllthcyBlockBlock 2Which of the two blocks will tip...

Question

Thesolid; uniformly dense blocks of wood shawn below are both slovtly tilted differcnt helghtsincressing angle untllthcyBlockBlock 2Which of the two blocks will tipover first? Block 1 will tipover at a smaller angle: Block 2 will tlp over at a smaller angle We cannot say which block will tipover first without knowing the exact measurements of the blocks Both blocks will tipovcr at the same angle:

Thesolid; uniformly dense blocks of wood shawn below are both slovtly tilted differcnt helghts incressing angle untllthcy Block Block 2 Which of the two blocks will tipover first? Block 1 will tipover at a smaller angle: Block 2 will tlp over at a smaller angle We cannot say which block will tipover first without knowing the exact measurements of the blocks Both blocks will tipovcr at the same angle:



Answers

A cubical block rests on an inclined board with two sides parallel to the incline. The coefficient of static friction between block and board is $0.95 .$ If the inclination angle of the board is increased, will the block first slide or first tip over?

In this question. We have a wood block sliding up a would ramp, and we're trying to find the smallest angle that will allow the block to fall back down after it reaches a tight point. Mhm. So we can start by making a picture of the situation. So we have the ramp at some angle theater and the block sitting on the ramp, and we need to find the angle that will allow the block to fall back down. So I'll draw the block falling down the ramp, and I'll set up the cordon system such that the X axis is along the surface of the ramp. Mhm so we can make an F B D. So we have gravitational force pointing downward. Normal for is perpendicular to the surface of the ramp and some frictional force opposing the motion of the block, which is going down the ramp so the friction must act up the ramp. And this friction will actually be the maximum static friction because in order for the block to fall back down the ramp, the max static friction must be overcome so we can look at you in second law in the X direction we have F net is equal to M A, which is also equal to zero because of the static friction and looking at our picture here we have the force of gravity pointing downwards and we can make a right triangle where this angle theta is the same as the data of the rent. Yeah, and we can split the force of gravity into its components. So along the x axis, we have F G sign data and along the Y axis, we have F G coastline data. So for F neck, we have negative f g sine theta. Plus the force of friction is equal to zero mhm where F g is equal to mg. So we have negative mg sine data plus use the best times the normal force which is equal to the magnitude of the max. Static friction is equal to zero and then we can add mg sine data to the other side. So we have a new service times The normal forest is equal to mg sine data and now we can look in the Y direction in order to solve for the normal force. So we have F net is equal to m A which is also equal to zero since the block doesn't move in the Y direction. So we have our normal force minus f g cosign theta is equal to zero. Yeah, and solving for the normal forest we get that the normal forest is equal to mg cosign data. Yeah, and now we can plug this back into our first equation. So we have use of s times mg coastline data is equal to mg sine data dividing by mg coastline data we get new suggest is equal to science data over cosign data which is equal to tangent data. So therefore data is equal to the arc tanne of mischievous. Yeah. And, um, use of s for wood on wood is equal to 0.5 so state as equal to the ark, 10 of 0.5, which is equal to approximately 27 degrees

So in the previous question we have placed the block on the horizontal floor and the force was applying at the inclination at the angle of tita with the horizontal. Okay, and now we have placed the block at the in plantain at an angle alpha. This is the alpha angle with the horizontal and horizontal forces being applied at an angle this tita, Okay, this angle theta visa inclinations. So this cheetah will be equal to this alpha angle from the geometry we can see here. So hence these two conditions are same because here is the floor horizontal and the forces applying at the angle theta. And here is the force horizontal and the inclination angle of this plane is theta. Okay, so both conditions are same. So hence the locking conditions for the This block will also be same. So we can write that 20 to that must be greater than or equal to one by mu as we have opted from the previous question. So this will be the answer for the question and option D. It will be the correct answer here, so this become the condition for the question. Okay, so this becomes the answer for the question.

This is a problem that deals with the center of mass. The center of MASH equation can generally be written something like this, where the numerator has the center of mass of each individual object times, its mass divided by the denominator, which is just the sum of all of the individual masses. This equation looks a lot more complicated than it actually is. So let me explain this by drawing the picture first and foremost, so we have a table here. We have a leg, there's the top of the table and we have a second leg. And on top of the table there are three blocks. They're kind of stacked off of the table. The first one has one quarter of its length off of the table. The second one has an additional quarter of its length off the table, and the third one has once again an additional quarter of its length off the table. So all of these blocks have a total length of L because they're the exact same block, and they also have the same mass of em. So it looks kind of confusing, but it's not actually that bad. The first thing we need to do is set our X equals zero point and we're going to set that right here at the edge of the table because that just makes the most sense with geometry. Now what we need to do is find the individual center of mass is for each of these objects because it's just literally a block. The center of mass is just going to be the very center of the object. So I'm drawing the X here to kind of show where they are. Now that we have that we just need all of these values. So what I'm going to be looking for is the distance between the center of Mass and the X equals zero point for every block. And that's what we plug in for the X I right here. So if we just start plugging things into the center of mass equation, we have in the denominator the some of the three Masses, which is just M plus M plus m. You can also write that as three m. But I wanted to be a bit clearer here, so if we start with the red block because this distance here it's l over four off of the edge of the table. That means that from the edge of the table to the center of Mass is just going to be once again another L over four. And that's because the center of mass for each of these blocks is just going to be at half of its length because that's how center of mass works. So we can write up here that we have The center of mass of the red block is l over four. We multiply that by its mass. We will now add the center of mass of the green block. And this one is actually easy because l half of the green block, the very center of the green block, is on the edge of the table. So we have zero times m, which is kind of irrelevant, but we'll write it out anyway. And then we do the same thing for this last block. Now, my drawing doesn't really indicate this, but if this length here is l fourth, this length here is another l fourth and we have a third l fourth here. That means we have three l over four until we get to the very edge of the table. And what that means is that from the edge of the table to the center of mass is just a distance of l divided by four. The main difference here, though, compared to our example in red, is that now we're on the left side of the X equals zero axis. And what that means is we're not actually going to have a positive value of a center of mass. This is going to be a negative center of mass, so it will be negative. L divided by four, multiplied by the mass. Let's just make this denominator a little bit larger. So when you plug all of this in your you're obviously going to have 3 a.m. in the denominator, the numerator, the zero times M is irrelevant. But we also have that the red number l four times m is exactly equal, but opposite to the blue number, which is just the minus. L divided by four times m. So if you have the same number subtracting itself, that just becomes zero, which is the final answer. The center of mass of the final configuration is just zero that if the center of mass is zero right on the edge of the table. This is a stable configuration and it won't fall. It's close to it. If we were a little bit further to the left, if we had a negative X value for the center of mass, then it wouldn't be stable. But here it is.


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