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Current time:0:00Total duration:8:44

Vectors word problem: tug of war

Video transcript

teams a b and c are playing a game of strength each team has attached a rope to a metal ring and is trying to pull the ring into their own area team areas shown below so let's see what that's so we see that over your team a is pulling on the ring in that direction team b is pulling on the ring in that direction team C is pulling on the ring in that direction they're each trying to get the ring into into their respective areas it's a three-way tug-of-war they tell us team a pulls with the force vector a of four times the unit vector I plus a zero J so this is four times there's no vertical component here so it's going straight in the positive I guess you could say X direction and we can see it right over there a is going straight in the positive x Direction is just a multiple of the unit vector I team B pulls with force vector negative 2i plus four J so it's to the left and up and we see that negative 2i plus four J then we have team C pulls with a force vector negative 3i minus 3 J so let's see negative 3i minus 3 J yep that seems consistent with the diagram forces are given in kilo Newtons all right which team will win the game so think about which team will win the game we've to think about which one is pulling with the most magnitude in their direction so let me get my little scratch pad out and let's think about the magnitudes of each of these vectors so let's the first think about the magnitude of vector a what is that going to be well it's 4 times the unit vector I it's going to have so it's just going to have a magnitude of for the unit vector I has a magnitude of 1 so its magnitude is just four times that its magnitude is equal to four and let's put the unit's there it's four kilonewtons now what about the magnitude of vector B the magnitude of vector B well we can use the Pythagorean theorem it's negative 2i plus plus 4 J so the magnitude squared is going to be negative 2 squared plus 4 squared which is really just coming out of the Pythagorean theorem so the magnitude of vector B is going to be equal to the square root of negative two squared plus four or one way to think about this is the Pythagorean theorem or the distance formula between the origin and that point right over there plus four squared that's four plus sixteen this is the square root of 20 and let's see this is definitely greater than 4 4 squared is just 16 this is someplace between 4 & 5 and let's see team C the magnitude of vector C same logic it's going to be the square root of negative 3 squared plus negative 3 squared which is equal to let's see this is square root of 18 right 9 plus 9 square root of 18 so square root of 20 is clearly larger than the square root of 18 and we already said this quart of 20 is larger is larger than 4 so team B is going to win which team will win the game team Team B and we can even let's just fill it out on our team B is going to win the game now the next thing they asked this is what is the magnitude of the team's combined forces acting on the ring well to figure that out we can just add up all of these vectors the the combined force so let's write this so the force let's say force combined let's this is a vector it's going to be the sum of vectors a plus vector B plus B plus C plus C which is going to be equal to 4 I plus 0 J plus or minus 2i minus 2i plus 4 J minus 3i minus 3i minus 3 J and so now we can just add up the i's and add up the J's so we have 4 I minus 2i minus 3i so 4 of something minus 2 you're going to have 2 left then you subtract another 3 you're going to have negative 1 of that so that's going to be equal to negative I could write negative 1 I or just negative I and then we have 0 JS + 4 J's - 3 J's so that's what 0 plus 4 minus 3 is just going to be 1 J or you can say just plus J so that's the resulting vector here is negative I plus J so it might look something like this so negative negative I plus AJ gets us a vector gets us a vector that looks like that and let me actually draw it a little bit bigger here so if I were to draw if I were to draw our coordinate axes that's my Y axis this is my x axis x axis it's negative 1 times I so let's say it looks like that that's negative I that's negative I right over there negative I let me write that that is negative I plus J plus J plus the J unit vector which might look something like this both of them have a magnitude of 1 so that is the J unit vector and the sum of those two is our combined force our combined force is going to be going it's going to be going into this direction so this is its magnitude so the length of it is its magnitude combined force and that goes in line with we're going to be going in that direction which takes us to the zone of B but they ask us what is the magnitude of the team's combined force well the magnitude of this side is 1 it's a unit vector where it's going in the left direction the magnitude of this side is 1 so the magnitude of this side we just use the Pythagorean theorem it's a square root of 1 squared plus 1 squared which is just the square root square root of 2 so the magnitude here what is the magnitude of the teen combined the team's combined forces its square root of 2 and what direction is the ring getting pulled and actually let me just fill out what they're what they've already asked us just so we can so what is the magnitude of the two teams combined force acting on the ring round to the nearest tenth so right square root of two square root of two is equal to and round to the nearest tenth one point four kilonewtons one point four kilonewtons and what direction is the ring getting pulled and they want to send radians round to the nearest tenth your answer should be between zero and two pi radians so that go back to our diagram we need to figure out we need to figure out what this angle is right over here we need to figure out what that angle is in radians and we know we know that this whole thing if we went from the positive x-axis all the way the negative x-axis that is PI radians and we also know what this angle right over here is how do we know that well I've just constructed a notice this is a right angle and this is an isosceles triangle isosceles right triangle it's going to be a 45-45-90 triangle but if you're thinking in terms of radians 45 degrees is PI over 4 so this is PI over 4 PI over 4 PI over 2 triangle same thing as 45 45 90 so this whole purple angle is PI this is PI over 4 you just subtract the PI over 4 from the PI to get what theta is so theta is going to be equal to PI is the same thing as 4 PI over 4 minus PI over 4 minus PI over 4 I just did that to get a common denominator which is equal to 3 PI over 4 3 PI over 4 radians so let's go back over here now they want us to round it to the nearest tenth so let's take 3 3 times pi divided by next to make sure you can see what I'm doing here so 3 times pi divided by 4 is equal to 2 point 3 5 6 round to the nearest tenth would be 2 point 4 radians so approximately 2.4 radians and we got it right