 In this video we provide the solution to question number 14, which is going to be a problem about vectors. We have a 16 pound weight that's lying on a sit-up bench at the gym. If the bench is inclined at an angle of 15 degrees, there are three forces acting on this weight as shown in the diagram provided here. So if we kind of think of the picture to help us out here, we have this incline and then there's this weight that's sitting on it, like so. So because of the incline here, the normal force is going to be acting at an angle, gravity pulls it down, and then there's friction that's holding it up as well. And so that's why these three forces come into play. Let's just look at the simplified diagram. Our three forces, of course, are the normal force N, which is perpendicular to the bench it's sitting on. There's the frictional force F, which is due, again, to friction. So it doesn't just slide off, and then W is the weight of the weight itself. If the weight does not move, then the sum of the three vectors is zero. That is, it's in a state of static equilibrium. Let's find the magnitude of N and the magnitude of F in this situation. We know that the weight has a magnitude of 16 pounds because that's given to us. So then we have to decide how to solve this. We have two approaches we could solve on this one. We could switch everything to algebraic form and then just kind of work through there. It's kind of nice, but there's some conversion that has to go on there. We could also set up a triangle. The advantage here is that since the normal force is perpendicular to the frictional force, because the frictional force will go in the direction of the bench, the normal force will be perpendicular to the bench, this would set up to form a right triangle, and right triangle trigonometry is not so bad. Oblique triangle trigonometry can be very challenging. So I think I could rearrange this thing to form a right triangle. So let's think about how things would move around. Let's have the weight vector be right here. Then the normal, the normal, I'm going to move to be over here. So we get something like this and then the frictional one would move to be down here like so. So we get this right here. This is our frictional force. My N got erased, so I'm going to draw that again. This is something like this. So this is our force diagram. Because it's an static equilibrium, it should form a polygon and since there's three of them, it'll form a triangle. The angle between the normal and the frictional force, that should be a right angle. So it is, that's a right angle as well. Let's make that look like a right angle. How does this measurement of 15 degrees come into play right here? Well, this is with respect to the horizontal. So let's think about that, for example. So if we were to draw a horizontal line right here, that's going to be 15 degrees. OK, but the weight vector, right, that's perpendicular. This is, this right here is a right angle as well. The weight vector is vertical, which is perpendicular to the horizontal. So that means that this angle right here inside of the triangle, that angle will be complementary to 15 degrees. All right. And so in particular, that's going to be 75 degrees, which then as this is a right triangle, the other angle here is going to be 15 degrees. And you probably couldn't also discover that using some type of corresponding or alternate here in angles argument as well. But we have this triangle, which is a 1575 90 triangle. So we need to find the sides here. We know that the hypotenuse is 16. We need to use a sine or cosine ratio to figure out the other ones. Let's do the frictional force first, right? So if we take with respect to which one do we want to do? We'll do the 15 degree angle because we have that one. So with respect to this angle right here, the frictional force is the opposite side. The hypotenuse is going to be the weight, and this should equal sign of the angle. We'll call it theta, right? So we'll call this angle here theta just for simplicity's sake. So then that tells us, of course, that if we divide by every time both sides by the weight, we're going to get the frictional force is equal to the weight of the weight, which is 16 pounds times sign of 15 degrees. For which we can use our calculator there. But sine of 15 degrees is one we've done before with exact values. You'll get the square root of six minus square root of two over four, like so. Four goes into 16 four times. So you get four times the square root of 16 minus the square root of two. This would be in pounds, like so. If you want an approximate value, you could approximate with your calculator. That's unnecessary here. Exact answers are perfectly fine. Then to find out the other side, to find the magnitude of the normal force, we could just use the Pythagorean theorem using 16 and this value right here. But as this value has some square roots, it might be easier just to use the cosine ratio. Notice that the normal the normal force versus the weight, this is going to equal cosine of theta again with respect to the 15 degrees. This is the adjacent side. So this is cosine here. This tells us that the the magnitude of the normal force will equal 16 times cosine of 15 degrees, which is the same thing as the same thing as sine of 75 degrees if you prefer. So you get 16 times the square root of six plus the square root of two over four. And that simplifies similarly. You're going to get four times the square root of six plus this time the square root of two pounds, like so. And this finds our magnitudes. You are allowed a calculator on this section of the test. So if you want to approximate your answers or you feel more comfortable doing so, that is acceptable here. But I'm just going to leave this as the exact answers as I end this video.