Draw each of the following vectors. Then find its x- and y-components. (c) F = (50.0 N, 36.9 degrees counterclockwise from the positive y-axis)

Video transcript

Hey, everyone in this problem, we're asked to sketch the acceleration vector A to calculate the components A X and A Y of a cart pulled by a horse accelerating at 2 m per second squared, 35 degrees clockwise from the positive X axis. So let's start with this sketch. OK. So we're just gonna draw out our axis. OK. So we have our xry axis in this acceleration vector we're told is 35 degrees clockwise from the positive X axis, right. What we're gonna do is we're gonna go from the X axis in the clockwise direction. We're gonna go 35 degrees. Hm. And our acceleration vector is going to run from the origin at this angle of 35 degrees. It works down into the right. This is our acceleration vector A and we know that it has a magnitude of 2 m per second squared. Mhm. All right. So we've sketched that acceleration vector A now. And what we're left with doing is calculating the components A X and A Y. So if we take a look at the sketch that we've just done, we can draw out those components. OK. So the A X component is gonna be the horizontal component. So it's just gonna lie along the X axis and point out to the right, the Y component is going to lie along the Y or parallel, sorry to the Y axis along the vertical and it's going to point downwards. OK. We have A Y, we've drawn A X in blue A Y in green. Now we have a 90 degree angle in our triangle and we have a right triangle. We know the hypotenuse, we know the angle. So we can use our trigonometric ratios to calculate these components A X and A Y. I'm just gonna do a little bit of simple triangle math and we're gonna have those components starting with the A Y value. OK. Now A Y, if we look at our angle is opposite to that angle, so we wanna relate the opposite side with the hypotenuse, we're gonna use the sign I recall that sign of an angle theta is equal to the opposite side divided by the hypotenuse. So in this case, sine of our angle 35 degrees is equal to that opposite side, which is A Y divided by the hypotenuse 2 m per second squared. And what we've mixed up here is that this should be sign of negative 35. OK? Because that angle is going clockwise and not the typical counterclockwise. Yes, we have a negative in there. And so A Y and we can multiply both sides by 2 m per second squared A Y is gonna be 2 m per second squared multiplied by sign of negative 35 degrees. And so we get the Y component is going to be negative 1.147 m per second squared. Hm. Now, if you forgot to do that negative in the inside of the sign with the angle, you would end up with A Y component that is positive. OK. We would wanna go back double check with our diagram and think well A Y is pointing downwards that should be negative. OK. And then kind of go back and figure out where we've made that mistake. OK. So just always double check that the value you're getting makes sense with the diagram you've drawn. All right. So we have our Y component. Now we're gonna flip over to the X component. We have cosine of beta. Now because we're gonna relate the adjacent side in the hypotenuse cosine of the recall is the adjacent side by the, by the hyponym substituting in our values cosine of negative 35 degrees is going to be the A X component divided by the hippo 2 m per second squared. Again, we multiply both sides by 2 m per second squared. We have 2 m per second squared multiplied by cosine of negative 35 degrees. Hm We work this out, we get that the A X component is gonna be about 1.6 3 8 3 m per second squared. OK. So now we have a sketch of our acceleration vector A and we've calculated the A X and A Y components of that vector. That's it for this one. Thanks everyone for watching. See you in the next video.