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Current time:0:00Total duration:11:54

AP.PHYS:

INT‑3.F (EU)

, INT‑3.F.1 (EK)

, INT‑3.F.1.1 (LO)

, INT‑3.F.1.2 (LO)

, INT‑3.F.2 (EK)

I don't know about you but torque problems used to give me anxiety and I think it was because I didn't really understand well what torque meant or how to find it so what I want to do in this video is show you how to find the torque there are conceptual ways and tricks to figure it out so I want to share those with you so that going forward we don't have to be so anxious when we're solving a torque problem and specifically the problems I got most anxious about were these problems where the force was at a weird angle so let's do this let's figure out how to find the torque from say this 10 Newton force exerted at this angle of 30 degrees now one of the first things you want to do when finding the torque is identify the axis the axis is the point about which the objects going to rotate so let's say it was told to us that in this problem the object whatever it is rotates around the center so the center here of the object would be the axis maybe this is a board with a nail through it or maybe this is a bird's-eye view of one of those fancy revolving glass doors at the nice restaurants and hotels regardless let's say the axis was in the very center and that's crucial to know because if a force is going to exert a torque it has to be applied at some point besides of the axis in other words if you go try to push open this revolving glass door at the very center nothing's going to happen because it's not going to rotate but the farther out you apply this force the more torque you will get for the amount of force that you're exerting so a force out here would apply much more torque than a force right here that's why the doorknobs are near the edge of the door it'd be really hard to open a door right near the hinge if you haven't tried it try it out it's really hard now that we've identified the axis we could figure out how much torque we're exerting now the first thing I might try to figure out the torque here is I just say all right torque I know what torque is torque is F times D or F times or you could call it an R but what's important to know is that this R represents the vector that points from the axis to the point where the force was applied so in this case that would represent from this axis right here to the point where the force was applied would be this this would be R note that R is not the entire radius necessarily and it's not the entire length of the object always from the axis to the point where the force is applied and technically this R is a vector you can think of it as a position vector but regardless it points from the axis to the point where the force is applied it doesn't point the other way the direction is not toward the axis the direction is always away from the axis to that point where the force happens to be applied to the object so let's give this a number let's say this happened to be 2 meters from this axis to the point where this 10 Newton's was applied now we can solve for this torque but you have to be careful a mistake I might have made is to say just well the force was 10 Newtons the R here is 2 meters so my torque should just be 20 right 2 times 10 but that's not right because this force is not representing the total force necessarily if you just write this formula for torque like this what you really mean is that this force is the perpendicular force to this R so only the perpendicular component of this force is going to exert a torque on the door the component parallel to the R doesn't exert any torque and that should make sense if I draw this out let me draw the components if I break this 10 Newtons up into a component that goes this way I'll call that F parallel because that force is parallel to our it runs the same direction as R does and I'll break it up into this component as well this perpendicular component and I'll call that F perpendicular because this component is perpendicular to this R vector only this perpendicular component is going to exert a torque and that should make sense torque is a force that causes something to start rotating or to change its rotation so the only component of this force of this 10 Newtons that is going to cause this door to rotate is this perpendicular component this is the way you push on a door to make it rotate you don't pull the axis this way if I came up and tried to open this revolving glass door by pushing that way you'd think I was crazy because that's not going to cause the door to rotate similarly trying to pull the door that way is not going to cause this door to rotate you need to exert a force perpendicular to this R vector in order to get the door to rotate in other words only perpendicular components of the force that is perpendicular to the R are going to exert a torque so it's only this component of the 10 Newtons that's going to contribute toward the torque and we can find that if this was 30 degrees this is an alternate interior angle I mean this is also 32 so these angles are identical based on geometry that means this component that's perpendicular I can write as well let's see it's the opposite side that side is opposite of this 30 degrees so I can say that it's going to be 10 Newtons the hypotenuse would be 10 Newton's times sine of 30 and 10 Newton's times sine of 30 degrees is 5 Newtons so finally I can say that the torque exerted on this door by this force of 10 Newtons at 30 degrees would be the perpendicular component which is 5 Newton's times how far that force was applied from the axis and that was 2 meters and I get that the torque here is going to be 10 Newton meters so this point I wouldn't blame you if you weren't like see this is why I hate to work I've got to remember that this 2 meters is from the axis to the point where the force is applied I've got to remember that I'm only supposed to take the perpendicular component and I'm supposed to remember that perpendicular means perpendicular to this R vector you might wonder is there an easier way to do this is there a formula that makes it so I don't have to carry so much cognitive load when I'm trying to solve these problems and there is since this force component is always the component that's perpendicular to the R we can take that into account when writing down the formula in other words the way you find that perpendicular component is by taking the magnitude of the total force the 10 Newtons and you multiply by sine of the angle between the R vector and the F vector that's what we did to get this 5 Newton's so why not just write down this formula explicitly in terms of the total force times sine theta that's going to be the perpendicular component and then multiplied by R so what this represents is this here this F sine theta is f perpendicular and then you multiply by r just like we always do now in most textbooks you'll see it written like this they take they like putting the sine theta at the end looks a little cleaner so if we do F times R times sine theta now we can just plug in the entire 10 Newton's in for the force the entire 2 meters in for the R and this theta would be the angle between the force and the R vector but that's crucial you've got to remember if you're going to use this formula instead of this formula you've got to remember that this angle here is always the angle but mean the force vector and the R vector which is the vector from the axis to the point where the force is applied which in this case was 30 degrees so sometimes it's not obvious how do you figure out the angle between F and R well first you identify the direction of F and the direction of AR the safest way to figure it out would be to imagine taking this F vector and just moving it so its tail is at the tail of the R vector and then you'd want to figure out okay how much angle is there between this F and this R well alternate interior angles again that makes 30 so that is the angle the angle between the F vector and the R vectors the angle we're looking for when we're solving for the torque exerted by a certain force let's use this formula let me take this formula let's take this we're going to use this to solve another example because the only way to get good at this and to not fear it is to practice a few let's take our new formula torque is F R sine theta let's say there was a force applied right here and let's say you were given these distances here and we want to figure out how much torque does this force of 20 Newtons apply if it's at this angle of 60 degrees so we'll use our formula we're going to use this F is the entire F now we don't have to break the F into components we can just say that it's the entire 20 Newton's of force the entire magnitude of the force times R but we got all these we got three different R's here which one do we use remember or is it defined to be from the axis which again is going to be in the middle to the point where the force was applied that's this way so the magnitude of R is one it's not three it's not four if you're given multiple numbers you have to be careful you have to select that vector that goes from the axis to the point where the force is applied so this is the magnitude of that vector which is one meter and then it's going to be sine of the angle between the force vector and the R vector so think about it force goes this way down and right R goes to the left the true angle between R and F would be we'd have to imagine moving F so that their tail the tail and we could say that F is going to point down and right R goes to the left the angle between them would be this much angle now we can find that in a variety of ways one thing we could do is imagine making a right triangle here if that's 60 degrees and this is 90 then this has to be 30 degrees since the interior angles of a triangle have to add up to 180 and if that's 30 and this is 90 then that angle has to be 120 so we can stick 120 degrees up here is the actual angle between the force vector and the R vector now if you miss that the reason we're saying 120 is because 90 plus 30 is 120 so if the angle between R and F is 90 degrees plus 30 degrees then it's going to be 120 degrees that's why we're putting 1 or 20 up here as the angle between R and F but you might be concerned you might be like that was a lot of work I don't have to do that I just want to take my F vector and determine what the angle is between F and R do we really have to imagine moving it and you don't so it's not that hard yes technically tail to tail is the way to determine the angle between two vectors but head the head gives you the same angle so I could have just looked at this angle here I knew that this was 60 and I knew 180 minus 60 gives me 120 so that's another way to figure out the angle you put into here so in other words you do not have to imagine moving this vector tail the tail if the vectors are butting heads if you have your F vector and your R vector butting heads just find the angle between the F vector and the R vector that way it'll still give you the same angle which is 120 degrees now you might wonder what if I totally screw up what if instead of putting in 120 I just throw in the 60 I mean that was the angle that was given what would happen then turns out you'd still get the right answer the torque formula is kind in this sense because even if I put in 60 sine of 60 is the same as sine of 120 that's not a coincidence here it's because this angle here this 120 between F and R is supplementary to the 60 degrees think about it this whole angle from this point all the way to there is 180 if this is 120 the 60 degrees would have to be the supplementary angle because these have to add up to 180 and the sine of supplementary angles gives you the same answer so if I plug in 60 degrees up here would still work so long story short even though we define the angle between vectors as the angle between them when their tail the tail you put them head to head that still gives you the same angle as tail the tail and since we're taking sine that angle we can use either of these supplementary angles to get the same answer so just stick your F vector next to your R vector find either of these angles you can use that in this torque formula and you'll get the right answer so recapping you can find the torque from a force by taking the perpendicular component of that force and multiplying by the magnitude of the r vector where this R vector is the vector the points from the axis to the point where the force is applied and by perpendicular we mean perpendicular to the R vector or you could use this formula where F would represent the entire magnitude of the force R would be the magnitude of the R vector and the theta in here represents the angle between the force and the R vector when they're head-to-head when their tail the tail or because sine of supplementary angles are equal you could also take the supplementary angle to that angle between F and R