Consider a 100-N weight suspended by two wires as shown in the accompanying figure. Find the mag

Consider a 100-N weight suspended by two wires as shown in...

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are used to relate the angles of a force vector to its horizontal and vertical components. This mathematical relationship is crucial for resolving the force vectors accurately.

Static Equilibrium

In mechanics, static equilibrium means that all the forces acting on a body sum to zero, ensuring the object is at rest. This principle is used to set up separate equations for the horizontal and vertical components of the forces, which ultimately helps in solving for unknown magnitudes and directions.

Vector Resolution

Vector resolution involves breaking down a force into its horizontal and vertical components using trigonometric functions. This process simplifies the analysis of forces by allowing each component to be examined individually, particularly when the forces are applied at angles.

Free-Body Diagrams

Free-body diagrams are schematic representations that isolate a body and depict all the forces acting on it. These diagrams are essential tools in setting up the equilibrium equations needed to calculate force magnitudes and their components.

Vector Addition

Vector addition is the process of combining multiple force components to determine the net force acting on a body. In a state of static equilibrium, the vector sum of all forces equals zero, which forms the basis for setting up the component equations.

Transcript

00:01 So if we want to try to find these force vectors, the first thing i would do is draw these right triangles like i'm doing here.

00:11 So both of those are going to be right triangles.

00:13 So this first one over here, notice we can find what is going to be like the x and the y components of this by doing the following.

00:25 So let me just write one over here.

00:26 Notice if we do cosine of 30 degrees to start, this should be equal to the angle adjacent to this.

00:36 So adjacent is going to be x, and then our other one for y.

00:48 Oh, actually, our hypotenuse almost forgot about that, would just be our magnitude of f1.

00:54 So if we want to solve for what x is, this is going to be x is equal to the magnitude of f times cosine of theta.

01:06 But one thing we need to keep in mind when we're doing this, at least with how i'm going to set up like this force diagram, is that anything to the left of here is going to be negative.

01:16 So we would actually not just have the absolute value here, but we would have the negative of this.

01:24 All right and now we can do something similar for y so we want to figure that out now we can go ahead and use sine of theta so it would be sine of 30 degrees which should be opposite which is y over hypotenuse magnitude of f1 and in this case we're going to want this to be positive because this force vector should be pointing upwards or actually, i mean, if we look at it going like this, you can see how it's pointing up as opposed to going to the left.

02:02 So, yeah, the y component should be positive.

02:05 So again, we just multiply that over, and we get that y is equal to the magnitude of f1 times sine of 30.

02:18 Actually, i don't know why i wrote cosine of theta here.

02:23 And actually, we can go ahead and figure out what these are because cosine of theta, should be root 3 over 2, and then sine of theta should just be 1 1�th.

02:36 So if we want to write f1 as a vector, so f1 should be equal to, so i'll put a vector hat over this.

02:46 So negative the magnitude of f1 times root 3 over 2, and then the magnitude of f1 times 1 1 1⁄ 1⁄2.

03:00 Now let's do the same thing for f2 over here.

03:05 So if we try to set these up in a similar fashion, we'll first, we're going to have cosine of 45 degrees, which is going to be adjacent.

03:15 So now this is our x, and then this is our y.

03:18 So these are going to be different x's and y's.

03:21 So the adjacent would be x.

03:23 This would be over the absolute value of, or not absolute value, our magnitude of f2.

03:29 But keep in mind that since we're heading to the right, this x value should be positive.

03:36 So this one is going to be positive unlike the other one we had.

03:40 Okay, so i'll just go ahead and rewrite this.

03:43 So cosine of 45 is root 2 over 2.

03:47 So if we want to get x by itself, we just multiply that over.

03:52 So it would be root 2 over 2 magnitude of f2.

03:57 And this should be equal to x.

03:59 That on the left side.

04:02 X is equal to this.

04:04 Now we can go ahead and do the same thing, but for y, so this is going to be cosine, or a cosine, of 45 degrees.

04:17 So this should be opposite over hypotenuse.

04:20 So the opposite is y.

04:21 The hypotenuse is f2.

04:26 So y over the magnitude of f2.

04:29 And once again, you can see how the y should be positive, since this is oriented upward.

04:35 Okay, so we have that and then again we just multiply over and this should be also root 2 over 2 so in this case, why should just be root 2 over 2 the magnitude of f2 and then actually just write what this force vector is right here so this should be equal to the absolute value or the magnitude of f2 root 2 over 2 and then this should be the absolute value, i don't know why i want to keep the absolute, say absolute magnitude of f2 times root 2 over 2.

05:15 So it ends up being the same.

05:19 Now, what we want is something that counteracts this weight here.

05:26 So if we were to think about it, this would be going straight down for the weight.

05:33 And so this is supposed to be 100 newton, so we don't need to multiply it by like acceleration or anything.

05:38 So if we were to think about it, this, should be, well, zero because it's not moving left to right, and they will just be negative 100.

05:48 So i'll call this w for our weight.

05:51 So let me go ahead and add that down here.

05:54 So w is equal to 0, negative 100.

05:58 Now, if we were to add all of these up, so just like add all of them up, this left side here should be equal to the zero vector.

06:13 And then the right side, well, we would just add each of these components.

06:21 So, actually, let me do it like this.

06:29 So i'll first do these in blue.

06:32 So if i add all those up, that would just be the magnitude of f2, root 2 over 2, minus the magnitude of f1, root 3 over 2.

06:45 And then for this other component here we would just add all of these up which would give us the magnitude of f2 root 2 over 2 and then plus the magnitude of f1 times 1 half so we have this and i almost forgot minus 100 okay so we have this now now this is zero vector if i actually rewrite it, it is just going to be 0, 0 like that...

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