 Hello, everyone. My name is Brad Langdale and I want to help you through some net force problems These are out of the textbook we use in physics 20 and the first ones on page 132. It's number one Alright, so we got two dogs a and b They're pulling a sled across a horizontal snowy surface dog a is going to apply a force of 200 newtons forward All right, let's make a free body diagram for this. So 200 newtons forward is our first force A dog B is going to apply force 150 newtons forwards. Okay, so I'm going to draw that in there as well and I am making my vectors tip to tail The force of friction on the sled is 60 newtons backwards Maybe I want to make that a little shorter because we're making these free body diagrams folks We got to make sure they're more or less relative to the magnitude. They should be and very important If it's going backwards that force needs to be a negative number Alright, so it says that we also have the driver trying to slow down by putting a force in the sled of a hundred newtons backwards So I'm going to put another vector in there 100 newtons backwards, you know a little shorter. Hopefully then some of those other ones came pretty good. Not bad Okay, so what's the net force so now that we've got our free body diagram worked out Which helps us get the directions down our net force statement for this first one is pretty easy It's a 1d situation and it's just a matter of totaling up the forces acting on the sled because at the end of the day That's what net force means it means total up all the forces you have And I think the key here is just making sure that if you have any forces going left that you are Making them negative numbers Sort of like this Okay, so if I total all those up, I'm going to get 190 newtons It's a positive number when I total it so that means it's going to be going to the right Okay, well, let's try another one here The next problem we're going to try out here is a tractor pull. We got four tractors connected by strong chains The load is at rest to start off with this thing. We're going to pull a&b pull with 500 newtons north Sorry newtons east and 4,000 newtons east. So let's draw that out. All right, so I'm going to say that amount is going to be 5,000 newtons Try to squeeze another zero in there and then another one a little shorter 4,000 newtons Now the next one here C&D we're going to go 4,500 west and 3,500 west so I don't know maybe that's about 3,500 West and maybe about 4,500 Newtons west as well. I'm going to put a negative sign on each of those to show they're going in the west direction Now the interesting thing about this one is it doesn't it says there's going to be friction There's a magnitude. So we don't know What direction but we know the size of the friction is a thousand newtons and the question is well, which way does the force of friction go I'll make the force of friction red So what I need to think about is if there is no friction, which way would this object move? Okay, so let's think our totals so far. Okay going in total to the right Or to the east we have 9,000 newtons of force And so far going in total to the west we have a little bit less We have in total eight negative 8,000 newtons So if there were no friction this object that the tractors are pulling would move to the right. Okay would move to the west Pardon me to the east But since we do have friction here friction is going to try to like Move against and and try to slow this object down keep it from moving So that means that force of friction I'm going to have going to the left Or the west in this question. That's my force of friction So Now if I was to think about what my net force is Okay, so my net force the total of all the forces acting here Well, you can probably just look at it from the numbers we've got up here I've got a thousand newtons to the right. I've got sorry 9,000 newtons to the right I've got a total of negative now 9,000 newtons to the left I have no overall net force So this object is going to stay at rest and that's what they're talking about in question b If the load is at rest will it start moving? No, even though this force is acting on it We are going to have exactly the same amount of force in either direction So according to Newton's first law the object at rest will stay in rest because there is no overall net force acting on it Okay, there's a couple of 1d situations Let's look at a 2d situations. This is a neat one The page 134 number one asks you to look at the picture they have From a previous question and then make some changes and solve it So you can go through and look at this example in the textbook on page 134 to see how they do it I'm going to show you a little bit of a different way to do it By adding these vectors using a little animation. So we're going to add together here Let's see what we're going to add we're going to add 60 newtons of force From these two people pulling on a canoe One of them is going to be an angle of 40 degrees kind of going up. So instead of 20 degrees in this picture We're going to do an example with 40 degrees and the other person's going to pull down at 20 degrees Oh, I guess that one stayed the same And we're going to figure out what the net force is Now in this case, we can't just go and add these two numbers together. It's not just 60 plus 60 is 120 We need to do this In a 2d analysis just like we did back in the first unit Where we're going to actually add these two vectors together using a ruler and protractor But it's really hard to do a ruler and protractor here Without you know actually being in person with you. So there's a cool little thing you can try on the FET website There's a vector addition animation and that might be kind of neat to use for this one here So I'm going to flip over to that Hopefully and I'm going to see if we can Check this out. So let's see just like we can draw out vectors tip to tail with a ruler and protractor to scale This animation lets us do a pretty decent job. It's not perfect, but a pretty decent job of doing the same thing So the first thing I'm going to do is I'm going to draw a vector and that vector is going to be Uh at 40 degrees. Okay, kind of up like this 40 degrees And I'm going to make it 60 newtons Except that, you know, just like in real life 60 doesn't fit onto this page. So I'm going to divide Uh and make it 20 newtons. I'm going to divide it by three Annoyingly, I can't get it exactly bang on Okay, so it's going to be a little bit. Is it going to be a little bit off 40 degrees and 20 I can get it as close as About that 40.9 degrees and 19.8. It's going to be okay Just like when we're drawing with ruler and protractor, we try to be as accurate as possible But we might be a little tiny bit off It'll be okay The next vector we're going to do is going to be going kind of down as we can see here Okay, like that 20 degrees From the horizontal and it's also going to be 60 newtons So let's see if I can put another vector in there here Okay, so I want to go down 20 degrees and I'm going to make it also 20 Newtons long or as close as I can get it Again, it might not be perfect, but I think that's going to be pretty close So those are my first two vectors now my resultant vector is going to be The tail to tail and tip to tip vector that goes right in here Just like we did back in the first unit That's the resultant and the resultant in this case is the same as the net force So let's see if we can find the resultant here. We've got this extra little vector I'm going to put it So that it's tail to tail. See if I can grab it come on you Okay, maybe I'll move his head first. Okay tip to tip And tail to tail There we go So I get an angle of about 10 degrees and 33.5 Is my Magnitude so if I go through and multiply that again by the scale factor Remember, I took all the 60 newtons I divided by three I'm going to times it by three to get about 100 It's 100.5 but pretty close So my resultant here my net force is about 100 newtons And It's at about 10 degrees in the rcs system So that's a lot like just adding vectors like we were doing in the past in uniday But now instead of finding the resultant we're finding something called the net force very similar There's another example here that I'm going to have you try and work through as well So give that one a shot you can draw it with a ruler and protractor if you have one at home You can go through and try this animation if you want it works kind of nicely as you saw But I'm going to go through one last one with you here There's a little bit of a trickier one But it's a good one for us to practice. So it says to minimize the environmental impact Of building a road through a forest a logger uses a team of horses. Uh, so I'll draw a really really poor horse Oh boy, this is not going as well as I hoped Okay, looks like a reindeer. That's fine. It's kind of a horse and it's pulling two logs So a log a and a log b And it's got a rope tied around it to pull these logs Uh, they give the mass of the logs. It says that a is 150 kilograms And it says that b is 250 kilograms. So I'm just going to label that on my diagram Uh, and then it says that the horses pull with a combined force of 2600 newtons. So I might call that like an applied force Force app stands for force applied of 2600 newtons And it says the force of friction On both of these Logs is 2400 newtons. So here's what I'm doing in my free body diagram. I'm saying that the force of friction Is negative 2400 newtons Now I'm not worrying about like normal force or force of gravity here because the motion is going to be horizontal It's not going to be vertical and the normal forces and force of gravities Which is cancelled out anyways if I add to them together. So I'm really just worried about the Horizontal forces here. I'm worried about the forces that are co-linear to one another All right. So now what we're going to do is find the acceleration of the logs. That's different We haven't found acceleration with one of these net force problems before but it's not that hard to do Let me show you how you do this so First things first, I'm just going to do a little net force calculation Okay, so I'm going to write net force Equals and I'm going to write down the forces that I have in this question So I have force of friction and I have applied force like that. That's my net force statement That's a place that we want to start off here. Okay It looks so important. I'm thinking I'm going to kind of highlight it You want to start off with your net force statement? Then what we're going to do is we're going to Put in place of net force the total of the masses in this question So all of the logs together, I guess are going to be 400 kilograms Okay, net force just like net force means total of the forces here. We're going to be Total of the masses And we're going to get here An acceleration So what I've done is I've replaced net force with mass times acceleration I'm basically just using Newton's second law Right f equals ma and replacing the force with ma in my net force statement Now I'll substitute in all my forces from the question All right, so I have negative 2400 newtons for the force of friction And 2600 newtons for the applied force So 200 newtons of net force On the right hand side of the equation 400 kilograms times acceleration on the right hand side of the equation or left hand side So when I divide both sides by 400, I get 0.5 00 Meters per second squared as the acceleration of the logs together And so that's not too bad to do. I mean aside from just taking the net force which we had here And dividing it by the mass Kind of the same as what we've been doing the last couple of problems Now part b is really interesting It says if the force of friction on the log a is 900 newtons calculate the force exerted by b on a This is a tricky one. So I'm going to make a new free body diagram for it. So here's log a It's 150 kilograms And it has some forces acting on it. Let's make a free body diagram think about what's pulling on this log Well, it has that applied force from the horses still Of 2600 newtons All right, and it has some friction on it It says the force of friction on just this log now not both like together Not together like a moment ago, but just the 150 kilogram log a is 900 newtons I'm going to make that force of friction going to the left because it's going to slow down that log Now there's also another force on that log There's the force that b is applying to a I'll call that force of b on a and I don't know what that is That's the unknown in this question So how am I going to figure out what this force is? I'm going to do it with a net force statement. So here we go Whoops net force Equals what do we have for forces? We have the applied force We have the force of friction We have the force of b On a which is our answer So you can have more than two forces together in a net force statement. No problem Now what am I going to put in place a net force? Well that same sort of idea is before mass times acceleration So the mass now since I'm only talking about log a is going to be 150 kilograms And the acceleration of log a we just figured out in the last question at 0.500 meters per second square It's going to have the same acceleration as it did a second ago that hasn't changed The applied force is 2,600 newtons The force of friction is negative 900 newtons And the force of b on a is what we're going to solve for So if I go through and solve here, I'm going to multiply the 150 and the 5.5 together I'm going to subtract from both sides 2600 and add 900 I get for the force of b on a To sig digs to work out to negative 1.63 Times 10 to the 3 newtons Now if you want to see maybe a little neater way of working that out You can also check out the key in the answers in the back of the booklet Or if you need some more help you can send me an email. So I hope that helped and For more information on these topics, please check out my website ldindustries.ca