Hey yeah, make your hand an open palm with your thumb out to the side, as if you were going to give someone a high-five. Align your 4 fingers with r, with your wrist at the tail of r. Then sweep your palm clockwise until your 4 fingers align with F. If you do this, your thumb would have been pointing into the screen. It's the only possible way to make the movement. The direction that your thumb is pointing relates to the "direction" of the moment vector (though that direction is in line with the axis of the rotation of the moment in the 2D plane). Check out this video on the right hand rule: www.engineer4free.com/4/right-hand-rule-for-the-vector-cross-product
In this video I define the positive directions/senses as follows: Horizontal forces are positive if they point to the right. Vertical forces are positive if they point up. Moments are positive if they have a counter-clockwise sense. You are truly free to define those however you want, and at the end of the day you will get the correct answer. If you start telling people down is positive though you will certainly confuse them. Its common to pick counterclockwise as the positive sense for moments but not necessary. Notice where I come to the answer of Moment about A = -2,366 Nm, that I write it as 2,366 Nm Clockwise. If I had picked clockwise as the positive sense, I still would have ended up with 2,366 Nm Clockwise at the end, just It would have shown as a positive value instead of a negative value. Basically, just pick your positive directions and senses at the beginning of the problem and stick to them until the end of the problem. As long as you don't make and math errors, you will come out with the right magnitude and direction/sense.
At the end of the cross product, why did you do the part with the [001] and why is the unit vector for moment in the Z direction? I thought this was only 2d motion so I'm a bit confused. Edit: I understand using the right hand rule, the cross product is in the negative Z direction, but I am just confused about how this relates to the moment. Does this mean the moment is in the Z direction instead of in the clockwise direction?
Hey, yeah, cross product must be applied to 3D vectors, so if we're working in . a 2D space, we need to add the third element of zero for the computation to work. Watch this video though: www.engineer4free.com/4/right-hand-rule-for-the-vector-cross-product I discuss the right hand rule and why 2D moments vector direction points into or out of the screen, while their rotational sense is in-plane, clockwise or counter clockwise
Yes. Probably easiest to do it with geometry. Plot the point and the line of action of the force. Find the magnitude of the force. Find the perpendicular distance between the line of action of the force and the point of interest. Moment = magnitude of force * perpendicular distance ... rearrange to have perpendicular distance = moment / magnitude of force. Do some geometry to finish it off. These linear algebra videos will help: find the length of a vector (magnitude of a force): ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-QPaonprd93g.html and find the distance from a point to a line: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8bQlwo31k2E.html . You might get lucky and the geometry will be really easy or obvious and you won't need to fuss with all the vector algebra.
Why does this work for any point along the line of force? Doesn't changing one of the vectors in the cross product drastically change the result of the calculation?
It's just a very cool property of this application. Plot out a few points that are on the line of action of the force then run the cross product calculation, you'll get the same answer for all of them!
Yeah, I’ve got a full list of the hardware and software that I use on engineer4free.com/tools . I can’t remember which drawing program I was using for this exact video, but you’ll be able to determine it based on the publish date. I think it was sketchbook
Hey, for the vector approach why is only the Fy negative and not the Fx because before, to get our answer using Ma = fd we had to have both Fx and Fy components negative. From what I learned, I know the J product is always negative when using cross product but Im assuming that isnt why you did that since the 2 for r is still positive. If you can explain I would greatly appreciate it.
Sounds like a stupid question but I struggle with trig and am not sure how you determine which function (sin,cos,tan) to use when trying to figure out moments. Any tips for how to easily identify the correct one would be greatly appreciated.
Hey, draw a right angle triangle over the diagram such that you know at least one length and angle. From the angle's perspective, identify which side is the hypotenuse, opposite, and adjacent. From there, you can determine which trig function will include the known length, and the unknown length you are looking for using SOH CAH TOA. Once you write the trig function with the known angle and one known side, you just need to rearrange it to solve for the unknown. See this image: imgur.com/a/YCT6I1a
It’s the way you’re typing it in, missing some brackets for proper BEDMAS order of math. The expression is to 500 * cos(30) = 433. Some calculators require you to enter 30 and then press the CO’s button! And that will return the closing of thirty which you then multiply by 500. Your calculator also needs to be set to degrees.
Hey yeah, both work but are just two equivalent ways of doing the same thing. If you take the cross product of two vectors, you get a third vector that is orthogonal to both of the first two. In the case of a 2D problem, the first two vectors span the plane that is the page, and the cross product will then give you a vector that's normal to the page (straight into or out of it). If your thinking about using M=f*d, you don'y write f or d in vector form, just the magnitudes of f and d. Now for a 2D problem like this, when you find the moment about some point, the moment is going to have a clockwise or counterclockwise sense. If you remember the right hand rule of moments though, it's that the the sense actually wraps around the direction that the vector quantity is pointing in. Watch this video for what I mean: www.engineer4free.com/4/right-hand-rule-for-the-vector-cross-product. So even though it feels like we're not using vector notation in M=f*d, we're actually doing the same analysis, and we can interpret M in vector form if we want to. Hope that makes sense. Sometimes M=f×r will be easier (usually is in 3D problems) and sometimes M=f*d is easier (usually is in 2D problems). Just pick which ever is easiest for the problem you are given =)
Moment = position vector × force vector, in that order. Often you will see the expression written as M = r × F. An interesting property of this expression is that r (the position vector) does not need to go from the point we are taking moment about to the point of application of the force, it can go from the point we are taking moment about to ANY point on the line of action of the force. Knowing that can sometimes simplify the work required.
Hey, no. There is no dot product being performed in this video. At 8:07 I say "because of scalar multiplication..." According to scalar multiplication, [0,0,-2366] = -2366*[0,0,1]. If you need a refresher on scalar multiplication, plz check out this tutorial: www.engineer4free.com/4/find-the-scalar-multiple-of-a-vector it is not the same thing as the dot product. So in this video, we find the moment vector by taking r x F, and get [0,0,-2366]. A vector tells us both a magnitude and a direction, and if you are familiar with your i,j,k vectors (refresher: www.engineer4free.com/4/using-i-j-k-unit-vectors-to-describe-vectors ), you'll be able to notice that the magnitude is 2366 Nm and the direction is in the negative k^ (read as "k hat") direction. What I did in this video was just use scalar multiplication to pull out the magnitude from the vector, so that we got the magnitude times a unit vector, and because the unit vector was [0,0,1], or "k^" it was meant to be more obvious about the vector direction. Now though, for a moment to have a vector "direction" is a pretty abstract concept thatmakes little sense in real life. The actual rotational sense of a moment is actually wrapping around the "vector direction" according to the right had rule (refresher: www.engineer4free.com/4/right-hand-rule-for-the-vector-cross-product ). So according to the right hand rule, when our direction is in the negative k^ direction (pointing towards the -z axis) which is "into the page" if we draw the typical x-y plane as it is in this video. So to come back to your question, we're not doing dot products. We're not multiplying by [0,0,1] just because, I just pulled out a scalar from [0,0,-2366] and was left with -2366[0,0,1]. However it turns out though that in a 2D problem like this video where your page is the x-y plane as normally drawn, moments will only form clockwise or counterclockwise in the x-y plane (the plane of the page), therefore having a "vector direction" of -k^ or +k^, respectively. Turns out that if your doing 2D moment problems with the cross product, you'll always end up getting a moment vector that is [0,0,A] where A can be positive or negative, and like I just mentioned, a positive value of A means the moment vector is pointing in the +k^ direction (in line with the positive z axis aka pointing out of the page), and the sense of the actual moment is counter clockwise, where as a negative value of A means the moment vector is pointing in the -k^ direction (in line with the negative z axis aka pointing into the page, and the sense of the actual moment is clockwise). That's a lot longer than I was expecting to write when I got started. Please watch the tutorials that I linked to, without a solid understanding of them none of this will make sense. Let me know if this cleared it up!
Ah ha, I think you are confused and thinking if the Force was applied in the "corner of the angle". I have been studying other videos for vector statics. In which the problem was set up with "tricks" where the force "point of application" is in line with the "point of moment". Thus, making one component(either x-dir or y-dir) multiply a "zero value" for the moment length to the "point of moment". Not the case in this example. Hope this helps.
Hmmm yeah it's tricky to show, without showing a real hand! 😅. Check out this video too which talks a little more about it: www.engineer4free.com/4/right-hand-rule-for-the-vector-cross-product
Extend the arrow for Fx and Fy infinitely in both directions. This represents the line of action of each of those forces. Draw a line that is perpendicular to Fx, that also passes through point A. The distance on that line between point A and the line of action of Fx is 2m. Draw a line that is perpendicular to Fy, that also passes through point A. The distance on that line between point A and the line of action of Fy is 6m.
Hey thanks for commenting. I break the force into its x and y components, then to find out what sense (clockwise or counter clockwise) the moments cause, I look at then one at a time. Just imagine that there is a pin at A, and then decide which way the object would rotate about A when you apply each component individually. I do this at 3:36 and 3:56 in the video and discover that both components would cause a clockwise rotation about A. Because I define counter clockwise as the positive sense, then both moments from the components shows up as a negative in the formula. Hope that helps!
