It goes in the same directionīut magnitude is one, that's why it's a unit vector. We're doing the principal root, the positive square root of one, which is just equal to one, Going to be equal to the square root, or I guess What's that going to be? Well nine plus 16 is 25 so it's going to be the square root of 25. It's going to be the, the square of three fifths is nine over 25 plus the square of fourįifths is 16 over 25. What's the magnitude of unit vector U? It's going to be the square root of the sum of the squares of Now of our unit vector? Let me write that hat a little bit, that hat got a little crooked. If the work done by a force vecFhati+hatj-8hatk along a givne vector in the xy-plane is 8 units and the magnitude of the given vector is 4sqrt(3) then the. In the same direction but the magnitude here is now Same thing as the ratio up here so we're going Going to have, one way to think about it the ratio between these two numbers is the exact ![]() The horizontal direction and four fifths in the vertical direction. To, the magnitude we already figured out is five so it's going to be three fifths in ![]() Once again we justĭivide by the magnitude, magnitude of our vector. It's three in the horizontal direction, four in the vertical direction. It's going to be equal toĮach of these components, for A we just divideīy the magnitude of A. Instead of this littleĪrrow when you put this hat this denotes that you'reĭealing with a unit vector, a vector with magnitude of one. To also make it clear it'sĪ unit vector and not just a normal vector I'm going ![]() On top I'm going to put - Actually just to not confuse ourselves let's call it U for a unit vector. Vector, I'll call it A but instead of putting an arrow Way of thinking about it, if we divide each of theseīy the magnitude of A then we can construct this unit vector. Of these components of our vector A by a fifth, or another What could we do? If we scale everything down by a fifth, if we were to multiply each Another way of thinking about it, let's say we wanted to figure out a vector that goes in theĮxact same direction but it has one fifth the magnitude, it only has a magnitude of one. Let's say we wanted toĬonstruct a unit vector that has the same direction as A but Right over here is five so I could say the magnitude You might have just recognized that this would be a three, four,įive right triangle. Square root of nine plus 16, square root of 25, or it's going to be equal to five. Squared plus four squared or this is going to be the This length is going toīe the square root of the sum of the squares What's its magnitude? The magnitude is just the length of this vector right over here and we can use the Pythagorean theorem to figure this out. In the horizontal direction we're going to go four in The magnitude of vector A, well this would just be the length of it. What else do we know about this? We could figure out A's magnitude, we can denote it like this. Vector, let's say the vector A, and in the horizontalĭirection for every three that it moves in the verticalĭirection it moves up four. A unit vector is justĪ vector that goes in a particular direction that Then calculate the work done by these forces.Ī 10-N force is applied to push a block across a friction free surface for a displacement of 5.0 m to the right.What I want to do in this video is explore the idea of a unit vector. For each case, indicate which force(s) are doing work upon the object. The following descriptions and their accompanying free-body diagrams show the forces acting upon an object. A free-body diagram is a diagram that depicts the type and the direction of all the forces acting upon an object. On many occasions, there is more than one force acting upon an object. Since F and d are in the same direction, the angle is 0 degrees.Ģ. The applied force must be 147 N since the 15-kg mass (F grav=147 N) is lifted at constant speed. The displacement is given in the problem statement. ![]() Thus, the angle between F and d is 30 degrees. It is shown that the force is 30 degrees above the horizontal. It is said that the displacement is rightward. The force and the displacement are given in theproblem statement. Since F and d are in the same direction,the angle is 0 degrees. It is said (or shown or implied) that the force and the displacement are both rightward. The force and the displacement are given in the problem statement.
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