welcome to a lecture online and here are some examples of how to work with torque in statics problems so static problems are where everything is motionless that means there's no snow net forces in any direction and that means there's also no net torques in any direction and that's very important when we do problems like this where we have to use the concept of torque to find the solution so here we have a vertical pole we have a beam attached to the pole that is able to pivot like this we have a cable keeping it from moving like this and now we have a mass hanging from the very end of that beam now the the problem here asked to find the tension in the cable in order to do that we have to pick a pivot point and I'm going to pick pick my pivot point right over there the reason why I pick it there is because I have no idea what the forces are on the beam on this part of the problem so that by putting a pivot point right there it eliminates any forces going through the pivot point as being non-material to the problem all right I'm going to label that pivot point number one and so I can say based upon this that the sum of all the torques around pivot point number one must add up to zero and so now what I have to do is identify all the torques related to this problem relative to that pivot point so using a different colored pen I can say that definitely I have a force going in this direction and that pen doesn't work very well anymore so let me use a different color I think my blue paint is still working let's try that there we go so where we have the mg the weight of this object going down like that and of course the distance from the line of action of the force to the pivot point the perpendicular distance right here could be considered d1 so this would be d1 which is the distance from the line of action of this force to the pivot point now the big beam has a big mass so the center mass would be right at the middle like this and we have a big mg which is the weight of the beam acting through the center mass of the beam and the distance from that line of action of that force to the pivot point I can that d2 and then finally we have a third force here which is a tension in this cable right here which pulls in this direction if it didn't the whole beam and this mass would just simply fall down so there's a four is going in this direction so this is the line of action of the force and this is the perpendicular distance from the line of action to the opponent right there and I could call this here distance d3 now notice that these two forces and big mg and small mg with causes beam to rotate in this direction the tension of the cable would cause the road the beam to rotate in that direction those were the only forces acting on the beam so by definition I'm going to call clockwise a positive torque and I'm going to call counterclockwise a negative torque that's completely arbitrary I know some textbooks do it the other way around that this is the way that I'm used to using it so and again it really that makes no difference as long as you're consistent within the problem so what we're going to do now is we're going to sum up all the torques and again by definition a torque is equal to the force acting on an object times the perpendicular distance from the line of action of the force to the pivot point so make sure we remember this definition of torque so let's go ahead and add up all the torques so first ork is the mg which is acting in this direction so since it would cause the beam to rotate in a in a clockwise direction if this was only force acting on the beam it's a positive torque so it's plus the force mg times the distance the perpendicular distance from the line of action to force to the pivot point which is d1 I like to just first do it visually and a layer on figure out what the d1 d2 and d3 are equal to then I have a second force which is acting this way again if this was only force acting on the beam move it caused it to rotate clockwise so it's a positive torque positive mg times the distance from the line of action of force to the pivot point which is the distance d2 right there and finally I have the torque I'm not the torque but the tension in the cable that would pull the beam in the opposite direction I'm going to call that a minus torque minus tension times the distance d3 now I noted that the beam had a mass of 200 kilograms the way that the end had the mass of 50 kilograms I did not say yet what the length of the beam was so just imagine that the length of the beam is equal to 4 meters all right because we need to know that maybe not we'll see all right so what is the tension on that string well here's our equation it's an equal to 0 so let's plug in what d1 d2 and d3 are equal to in this particular case so 0 is equal to mg times D 1 and D 1 is equal to the length of the beam times L plus big mg times D 2 which is half the length of the beam which is L over 2 and on – attention times D 3 now what is d 3 equal to its distance right here now this distance here makes up H a right triangle here mix up this triangle right here so let me draw this triangle out here on the side so it's this is distance 3 V 3 then we have the part of the cable coming down here and this is the length of the beam this is L and that's the right angle and this angle here is Theta and theta was equal to 30 degrees now notice let me get rid of this piece right there notice that D 3 is opposite of the angle this is the hypotenuse of this triangle so I can say that D 3 is equal to the hypotenuse L times the sine of the angle theta because it's opposite to the angle theta so therefore the sine of theta will define V 3 so I can then plug in what D 3 is equal to it is equal to L times the sine of theta now since this whole thing is it equal to 0 notice that can actually divide both sides the equation by L we didn't even need to know the length of the of the beam so L L and L simply cancels out because the whole equation has an L in each term and it's set equal to 0 the next I'm going to do is move the component that has the variable and looking for which is the tension here to the left side of the equation so we have T times the sine of theta it's not equal to mg plus 1/2 times big mg and finally I'm going to divide both side of the equation by the sine of theta so that the sign of theta here cancels out and I have the tension here is equal to little mg plus 1/2 times the big mg all divided by the sine of theta and now I can go ahead and plug in the numbers so tension is equal to little mg little mg is 50 kilograms times 9.8 meters per second square plus 1/2 big mg Big M is the mass of the beam which is 200 kilograms times 9.8 meters per second squared and the whole thing divided by the sine of 30 degrees and the sine of 30 degrees I believe is 1/2 so where's my calculator there we go so we have 50 times 9.8 plus 1/2 a 200 is 100 times 9.8 and we divide that by sine of 30 just 1/2 and the tension in that cable is equal to two thousand nine hundred and forty Newtons and there you go that's how you do a torque problem igap this was confusing I'll come up with a few more examples for you to take a look at at least that's a very good straightforward example and after a while I think you get the hang of it of how to go ahead and solve these kinds of problems so let me do a few more examples

could you pls distinguish between the tow cases: calculating resultant and applaying static equilibruim.

i m frommorocco, my english is not good, but i learned that before applaying static equlibrium we must extact our system.

I think to be clear, you chould say: We extract the beam and we will make static equlibrium of it. so we konw wha is interior to the system and what is exterior. what do you think about this.

6:59

My next rapper name, Lil mg

You are a good teacher but in my opinion you elaborate too much on the algebra. At this level your audience should certainly be able to solve a simple equation without you having to show all those steps.

I believe that it should be positive when the direction is counterclockwise and negative when clockwise? I think what he is saying is the opposite

WOW this is amazing!! Thank you so much! You made this so simple!!! I wish all physics professors taught like you everything makes sense now !! Subscribed š

Sir what's the hinge force at A when it is vertical and horizontal

Thank You for simplifying it and writing out the equations step by step. I've really been struggling with this but you are an excellent teacher and have helped me so much!

Thank you very much for this video! Iām french so Iām training static and English at the same time š the perfect deal

Thanks! You are a great teacher!!!!

Quick question: can we just draw the d3 along the bar? The only component of tension that causes torque is the y component anyway, and this component is normal to the bar. I find it easier to conceptualize this way.

This was great, thanks!

I cannot express how good of a teacher you are!!!!

Alright I'm one step closer to passing PHYS 106. Thanks!

Thank you

Legend. So Helpful

thank god for you

my left ear learned the sound of static and effectiveness of learning with a left ear on glitched video

Why we didn't get also the tension in the Y direction by making the tension = Tsin(30), the whole equation will be (Tsin30 X Lsin30) = (MgL/2) + (mgL)

Please answer me.

Thanks.

There are 4 forces acting on the beam (not 3 as stated). Weight of beam, weight at end, tension and normal force at hinge.

You just made the concept really simple, thanks.

Do we not consider the force exerted by the wall at the pivot point as well? My book has a similar example and that is what they did.

In previous video you said clockwise is the negative direction, why you chose the directions reverse ?