In the 3D spatial problem below, force F is represented by...

In the 3D spatial problem below, force F is represented by points DG and has a magnitude of 55.0 lbs on the end of the pipe segment as shown. a. Define the position vector AD. b. Define the position vector DG c. Define the unit vector DG. d. Define the Cartesian Force vector for the force F of 55.0 lbs. e. Calculate the direction cosine angles ($\alpha$,$\beta$,and $\gamma$) of the force F. f. Determine the moment produced by the force F about the origin, A. Express as a Cartesian Moment Vector & magnitude. (use the position vector from A to D.) g. What is the perpendicular distance between the point A and the force vector? h. Determine the moment produced by force F about wire AC of the pipe assembly. Express the result as a Cartesian vector and as a magnitude. i. What is the angle between the force vector and the wire AC? $\vec{r}_{DG}$ = {-6$\hat{i}$+7$\hat{j}$+6$\hat{k}$} 11 ft a) $\vec{r}_{AD}$ = {3$\hat{i}$+4$\hat{j}$-12$\hat{k}$} 13 ft $\vec{U}_{DG}$ = $\vec{F}$ = {-$\frac{6}{11}$(55)}$\hat{i}$ + $\frac{7}{11}$(55)}$\hat{j}$ + $\frac{6}{11}$(55)}$\hat{k}$ = {-30$\hat{i}$+35$\hat{j}$+30$\hat{k}$} 55 lbs M = Fd d1 = M/F $\vec{M}_{Axis}$ = $\vec{U}_{Axis}$ x $\vec{M}$

Transcript

00:01 Problem we have our circular beam with a point o and then some force f applied at the free end and we know that the from the center of the circle the force is applied at a 45 degree angle and at the radius of this arc 3 .55 feet and the force f 620 pounds so we want to find the moment that force f creates about 0.

00:34 This we know the moment formula where moment is equal to the force times the distance where the distance is perpendicular.

00:43 So to do this we can extend out our line of action of force f and then draw a perpendicular line from that line 2 .0 and this distance here distance d is what we want to use for this formula.

01:00 So we can work this out with some geometry if we have the floor because this is the center of the circle.

01:08 We know that this half is also the radius, so 3 .55 feet.

01:16 And then we know that this angle, because it's two intersecting lines, the angle across like this will be the same, so 45 degrees.

01:26 We'll draw this off to the side for a little more conventional view.

01:30 Our angle, the hypotenuse is 3 .55 feet.

01:35 And we want to find distance d.

01:38 So to do this, we'll use trigonometry and specifically that the sign of theta is equal to the opposite over the adjacent.

01:48 Here it'll be the distance over the radius because we have our opposite and then the adjc.

01:54 And this is not the adjacent.

01:57 My apologies.

01:58 It's the hypotenuse.

02:00 So the long edge of the triangle, which is the 3 .55 feet.

02:05 So we can solve this for distance and plug in.

02:10 We have a radius 3 .55.

02:14 And then the sign of 45 degrees, we put that in a calculator to get 2 .51 feet.

02:25 So we also should look at which sign moment will have.

02:31 If we put a conventional axis on this page, we'd have x and y.

02:38 So using the right hand rule, if we had our fingers pointed in the x direction, our palm in the y direction, our thumb will be pointed out of the page.

02:51 So we'd have our y -axis something like or z -axis something like this and we use a right -hand to do this so this is a right -handed coordinate system now this moment which is in the x -y plane is the moment vector would also be in the z -axis and so depending on its direction will depend on whether it's positive z or negative so because we have this line of action we can extend we can apply the four at any point along the line of action and it's equivalent.

03:33 So let's say that the force is applied right at the end here.

03:37 Imagine point o is a pin and force f tends to rotate around point o at the same distance.

03:44 We see that this is a counterclockwise rotation.

03:51 So if we had our fingers rotating in this direction, our thumb would be pointed out of the page, and that is the direction the moment vector goes.

04:01 So our moment is out of the page, the same direction as positive z.

04:05 So it is a positive moment.

04:08 This is basically arbitrary, but we usually use right -hand coordinate systems.

04:13 And so to keep this right -handed, we need to have positive moment be positive z in this way.

04:19 But we can just plug in this formula to get the numbers, the magnitude of the moment.

04:25 We know the force and the distance that we just found, 2 .5 .1 feet.

04:29 And so the moment is equal to positive 1 ,560 pound feet, or foot pounds, probably more typical, based to three sig figures...

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