😊😊 You should check out engineer4free.com/statics for all of the ither videos I did too!

Im lucky that my teachers are good. I just forgot this and cbf loggin in to find the tute. Vids good

Literally same here

🙌🙌 Thanks Korajark!! If you haven't already, do check out the whole collection of statics vids I made over at engineer4free.com/statics and tell some friends!! =)

Glad it helped! Check out the rest of the playlist here: engineer4free.com/statics =)

Glad I can help!! There are more vids here: engineer4free.com/statics =)

Thanks for the feedback Kammy, glad the videos are working =) =)

Hey sorry this comment wen't without a reply for so long, somehow it slipped through. Thanks a lot for your feedback!! :D

Thanks for the nice comment, and glad I can give you hope. please share the vids with other students that could use them too

Thanks Johan, It means a lot to me just to get a simple thank you but I always really appreciate a comment with genuine substance :)

Hey awesome thanks for writing Peter. Yeah I figured no one needs to sit and watch me write all that stuff out in real time =) =), glad you like the way I do it!

Glad to hear it bro. Just tell some classmate that could use the help too :)

You are very welcome!!

Keep it up and good luck!!!

Glad it helped!!! There are some more examples in videos 42-51 here: engineer4free.com/statics

Glad to hear I could help!!! It doesn't need to be complicated =) =)

@@Engineer4Free didn't even extend it till 10:01 for midroll ads, and for that I am eternally grateful

Thanks Bryce

Glad I could help! Full playlist is here: engineer4free.com/statics including lots on trusses

You're very welcome!! Incase you haven't seen the full playlist yet, it's here: engineer4free.com/statics plz share with some other students 🙏

@@Engineer4Free Definetely !!!!!!!

Thanks Nick, glad I can help.. you can make it through Statics!!! 🤜🤛

Thanks Nick, glad I can help.. you can make it through Statics!!! 🤜🤛

Thanks a lot! I hope you get lots of use out of it, and feel free to tell some friends so they can benefit too =)

Yeah true. If you can see that, then you're doing alright =) =)

@@Engineer4Free could you please explain what ou mean? if you take moment about point f wouldnt the forces we want to find be acting in the point and not formed into an equation

@@nahilismail3144 the force EG isn't, its perpendicular distance from point F is 0.866m- so you can calculate EG and sub it into Fx equation 🙃

@@inyoungbaek6426 thank you so much for helping a stranger you know nothing about. i hope you have a good day!

Yeah, I was thinking the same. But finding out all the unknowns can be useful for checking the answer by summing all the moments about an arbitrary point and see if that equals to zero.

Glad I'm able to help, you can see all of my statics videos here: engineer4free.com/statics there are quite a few truss videos!

Thanks Abdullah, glad I can help ✌️

Thanks dowg.

Thanks for the contribution Toan! It's a good sign if you are recognizing other ways to solve problems, shows that you are really understanding it. Thanks for contributing the discussion and showing others!

Good I got the same answer!!

Thanks Munashe =)

hah thanks!

Thanks cielcloud!! 🙌 You should also check out the rest of the playlist here: kzfaq.info/sun/PLOAuB8dR35oeXMk2C5fjHP2K306hGfk_w

@@Engineer4Free I already have your website bookmarked :D

Glad it was helpful!! =)

Glad I could help!! Check out all the statics vids I did over at engineer4free.com/statics when you have a minute ✌️

Thanks Dmitry!!

Typical haha. Lemme know how it goes!

Glad you liked it =)

Thanks Muhammad!!!!

You're welcome, thanks for watching! Check out the rest of my statics videos here if you haven't already: engineer4free.com/statics =)

😂😂.... But actually 🤔

One day we'll get to this point.

You're welcome Paul! 🙂

You’re welcome!!

Yeah so in the first line of the sum of moments about A equation, notice that there are two terms that include FG. One has an FGsin60 and the other has an FGcos60. The FGsin60 is the y component of FG, and it's also multiplied to 2.5, which is the perpendicular distance from A to the line of action of the force FGsin60. The FGcos60 is the x component of FG, and it's also multiplied to 0.866, which is the perpendicular distance from A to the line of action of the force FGcos60. Notice that the line of action of the x component of FG (FGcos60) is a horizontal line, in line with the top of the truss, which is 0.866m above the bottom of the truss. Both x and y components of FG cause a moment about point A, so both need to be accounted for in that equation.

You’re welcome!!! 🙂

You're welcome Lekhraj!! 🤜🤛

It's because Ax is can be determined to be equal to zero by inspection. There are no horizontally applied loads and the structure is simply supported, so Ax must = 0. Because its zero, I just don't bother writing it in the Fx expression. Hope that makes sense!

ah no, it should be 1m*sin60, as each member is 1m long =)

Hope you're exam went well!!! 🙌🙌

same

Because the line of action of EG passes through A. a force can only cause a moment about a point if it's line of action does not pass through that point

Ahh not quite. You got the first part right. If a member is pulling on a joint, then that member is in tension. Think about rope. Ropes can only be in tension. If you have a rope with a joint attached at each end, and that rope has been tightened somehow such that it is pulling on each joint, then the rope is in tension. The rope that pulls on support reactions is not in compression, and nether would a steel or wood member that is similarly "pulling" on each joint. If you're still not convinced, switch the perspective that you observing from. We're talking about the member pulling or pushing on the joint; so this what's going on from the joint's perspective. Try thinking from the member's perspective. If a member in static equilibrium is pulling on a joint, then the joint can be thought of as pulling on the member, just in the opposite direction. So if you're a two force truss member, and a joint is pulling on each side of you, then you're getting stretched out. Getting stretched out mean's that you're in tension. Conversely, if you're a two force truss member and your pushing on each joint, then those joints are pushing back on you. You're getting pushed on both ends. You're getting squished. Getting squished means that you're in compression. Hope that helps!

Ok so you are all good? Can't tell if the text after the edit is still relevant 🤔

There are no externally applied forces with any horizontal component. Ax is the only reaction force that can withstand horizontal loads, so the sum of forces in x direction equations is just Ax = 0. I skipped it, because it's almost always zero in these problems. So you could label it on, but it equals zero anyways and has no affect.

You could still solve the problem if you took sum of moment about F. Sum of moment about F would give you EG. Sum of force in y would still give you FG. You would find FGx as usual, and then use it, and EG, in your sum of force in x to find the last unknown, FH.

FG is diagonal so has both x and y conponent. Both of their lines of action do not pass through point A, so they will cause a moment about A if their magnitude is not zero. That's why we use both. The line of action of EG passes through A so it cannot create a moment about A, so it just gets omitted from the equation.

🙌🙌

You’re welcome Omar! 😁

Either works. If doing method of sections, make the virtual cut through members JH, JI, JK, and if doing method of joints, find the reactoins, then solve joint k first. Next you would solve for joint J and would find the answer.

You're welcome!

In this example, you have FH = -3kN then plugged in the Sum. Fy equation to get have a value of EG = +2.885kN. What If I plug it with +3kN since I know that FH is in tension and changed the graph. The value is different I guess.

​@@shariffnurrmohammad1314 Hey yeah. So just to clarify, FH is actually in compression. I drew it initially in tension, and found it to have a negative magnitude, which means it's true sense is compression, not tension. Once you know that a member is truly in compression you can do two things. Method #1) Do not update the arrow on any FBD (ie, keep it drawn in the tension direction) and then any time you analyze a FBD that involves that force, use it's direction to determine whether you add or subtract in the sum of force equation, but then substitute a negative value for the letter. That's what I did in this case. Calculated FH to be 3kN Compression in the moment eq, and then in the sum of force in x equation, I left the arrow for FH as if it were in tension, and because FH points to the right on this FBD, I give the FH a + sign in front because it points in the positive x direction. But then irrelevant to that + sign, I substitute "FH" for "-3kN" and "+FH" becomes "+(-3kN)" which is equivalent to "-3kN". Method #2) Update the arrow on all FBD's that it interacts with to show that it "pushes" on the joint rather than "pulling" and assign it a positive value. So if we did this, then we'd have to update FH at joint F to push on it (to the left) rather than pull (to the right). When we take the sum of force in x direction, then we'd give the FH a - sign in front because it points in the negative x direction. So we would have a "-FH" already in the expression, and then we would substitute "FH" for "(positive)3kN" and "-FH" would become "-(3kN)" which is equivalent to "-3kN". Hope that helps to clarify. We in a way have two layers of +/- rules going on here. One is that when solving for the unknown internal force of a member assumed to be in tension, that + means it is in tension and - means that it is in compression, and the other sort of unrelated thing when we are taking sum of force in x or y direction, that + refers to whether a force in a FBD is pointing toward the positive direction, and - refers to a force that is pointing toward the negative direction.

By drawing. right angle triangle that includes FG, FGx, and FGy and applying SOH CAH TOA. All angles are 60 degrees. See this: i.imgur.com/dTR8W7X.jpg

Hey. If you are unsure, of which to use, draw a right angle triangle overtop of the truss that includes the angle you're working with, and two of the sides along the members that make up that angle. From there, label the angle, opposite, adjacent, and hypotenuse sides. From there you can determine whether you need to use sin or cos. See this image: i.imgur.com/Dvdqf2t.jpg

When in doubt, draw a right angle triangle over top of the truss, using the angle in question, and both members that touch it as 2 of the 3 sides of the triangle. Then label on the opposite, adjacent, and hypotenuse sides, and from there you can determine if you need to use sin or cos to solve the unknown in question. See this image: i.imgur.com/Dvdqf2t.jpg

Hey, the equation for moment is M=Fd, where MN is the moment caused about the point, F is the magnitude of the force, and d is the perpendicular distance from the line of action of the force to the point. 2.5m is the perpendicular distance to the line of action of the y component of FG only. 0.866m is the perpendicular distance to the line of action of the x component of the force. Neither 2.5m or 0.866m are the perpendicular distances to the line of action of the force FG. It’s longer to try to find out what that distance is, than it is to just separate it into x and y components and solve by parts.

Hey I just have a few friction problems here: engineer4free.com/statics that are at the basic level of statics. I'll be getting around to friction problems in dynamics soon, and eventually may cover angle of repose if I ever do a serious on geotechnical engineering. Thanks for the suggestions =)

Thanks Penh 🤜🤛

Yes. You would need to include it in the FBD of the whole structure, when calculating the reactions at A and B. It will cause A and B to be different than they are now. After that, proceed in the exact same way. You can make the same cut, and inspect the left side of the structure if you want, but you will get different answers, because Ay will be different than it is now.

@@Engineer4Free but with the problem I’m doing, it asks my to find EG but joint G has a downward force acting on it. But when I split the truss in half, it gets cut off. What do I do?

@@SHIZZLE71 That's ok, it's not a problem. Go to 1:40 where I take the FBD of the whole structure. Include your extra downward force, at G. So you'll have an extra term in the sum of y expression, and also the sum about A expression. You will fin that Ay will be bigger than if your downward force did not exist. Once you found Ay, then do the cut, and analyze the left side. The downward force at G will not be present, because it is on the left side of the cut. But the influence of that force is already accounted for. This is why Ay is bigger than if that force at G was not there. It's because Ay knows its there from the calcs at 1:40. When you solve for FG, FH, and EG, their magnitudes depend on Ay. And Ay depends on all of the external forces, including the one at G. Hope that helps.. If it's still not making sense, then I just recommend watching vides 42-51 here: engineer4free.com/statics After doing several examples you might start noticing the pattern that they all follow, especially that Ay and By depend on the location and magnitude of the externally applied forces.

If you're cutting the truss in half to completely separate one side from the other, then make sure to only pass your virtual cut through a maximum of 3 individual members. If you for some reason pass through 4 or more, you won't have enough equations to work with.

tnx for your videos 😁

The angle might have been different on your test, or you may have needed to use cos instead because of the certain context. You have to take care when constructing the right angle triangle that you will use for your trig functions, here is an example of how to do one,: imgur.com/a/YCT6I1a sin is usually used for the vertical component, and cos is usually used for the horizontal component, but not always, it depends how you draw the right angle triangle

Sum of forces for the entire structure shows us that Ax = 0, so I just don't bother writing it

If you are talking about when he is doing the moments about A (time 3:30 ) then the answer is cause EG is pulling straight in the X direction with both Ax and EG having the same Y coordinates. If you are talking about the horizontal forces summing, he does have EG in there. If you are talking about the vertical forces summing, it is not in there because EG is *not* a vertical force.

@@joshman9757 Yup thanks dude. The line of action of EG passes through point A, so it doesn't cause a moment about point A =) =)

All members are the same length, so it's an equilateral triangle. All angles of an equilateral triangle are 60° (180/3). That is something you may be expected to realize on an exam so watch for it!

It's because the line of action of EG passes through A, so EG does not cause a moment about A. No need to write a term for a force in the moment about A eq if it's known to not be a part of it anyways 🙂

@@Engineer4Free Ah I see, thank you so much! You really helped me out :)

As in forces in the x direction

Because there are no horizontally applied forces on this structure. So if you wrote the expression for sum of forces in x direction, you would just have Ax = 0. Ax is the only thing that can apply a horizontal load to the overall structure (ie, not an internal force), so it just equals zeor. Because of that, I opted not to write the expression. If you're in a test, it's worth writing, but after making so many of these videos I started leaving it out. The rest of the playlist is here btw: engineer4free.com/statics =)