Visit ilectureonline.com for more math and science lectures! In this first of the seven part series I will show you how to find the tension of a cable attached to a wall and rod with a mass hanging at the end of the rod.
Hi Michael, been taking physics since grade 11 (I'm 25 now and am in first year engineering, sad I know) and for the first time I actually am feeling like I got what it takes to get good at this. I try to do these problems by myself before watching through and thanks to your excellent teaching methods I am finding at least half of the time I am on the right track... much better than before!
Don't worry about the "late" start on your engineering studies. (I didn't complete my bachelor degree until I was 27 years old, before I started graduate school). Life happens and we have to live it as it comes. We are glad the videos are helpful.
@@MichelvanBiezen yeah I wound up working trades for five-ish years before I decided I wanted to be an engineer, I didn't try my best in high school so I had to upgrade practically all my grade 12 credits to even qualify. But now that I'm here I don't wanna give up! Happy to report that I did pass my Physics Mechanics course, I imagine you're a big part of that! So thank you for your encouragement, here's hoping Waves and Fields will go well.
As a physics student, I just wanted to say thanks for all your videos! Even if they are not exactly what I'm looking for, they are quick and to the point which is perfect and helps me solve the problem at hand! Thanks again!
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!
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 :)
Excellent example. This is my "go to" video for tougher problems. I don't like the way you did the CW and CCW torques (I'm used to CCW = + and CW = -), but eh.....it works out in the end if you stay consistent with the torque positive/negative directions.
Anne To find the horizontal force on the hinge, you must sum all the forces in the x-direction. Therefore Fx = T cos (30 degrees) To find the vertical force at the hinge, you must use the sum of the torques with the pivot point at the right end of the horizontal beam.
By using this relationship can we make an easy-to-use hand pumping system for hand pump Now we have to make much effort to pullout the water from Borewell ?
Wow, you are such a great teacher! You put this wayyy better than my professor. Are you a Professor at a University? Because you should be!! Saw your Cal videos and your Physics tutorials are great as well, so far the best on RU-vid in breaking down these concepts. Very clear and straightforward. Do you do Chemistry tutorials as well?
Wow!, thank you so much, your explanation is so straight forward and understandable, you're super cool. My prof never explain where he got anything that he writes, he just flows while we are not flowing with him.thank you so much!!!!!!
What if the beam is not level and the big force is actually on a ramp connected to the end of the beam? I have a bus with a foldimg rear door(5ft.) that is part of the ramp. The ramp extends another 7 ft to the ground. The pivot is 40" high. The door would need to be lowered to 25" to make the ramp accessible to the car, then lifted once the car is pulled into the ramp to straighten the door and ramp combination.
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.
What if (for the CCW moment), we multiplied the component of T (Tcos30) with the perp distance (Tsin30 OR 4tan30)? I tried but I got a different answer and I do not know whether it's due to a calculation error or a logical error. Could you help me verify the kind of the error by double-checking my method?
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.
in the beginning of the video, is it possible for the stick to stay in balence if there is no string to support it? and lets say that the stick is fixed to the wall
What is material and diameter of cable to withstand the tension, also length and height of the wall and mounting of the cable is going to matter in the above example
At 3:41, you mention how turning clockwise gives you a positive torque, but using the Right hand rule, doesn't it state that a positive torque comes from Counterclockwise rotation giving you a negative value rather than positive?
My professor did a similar problem to this one and he multiplied the w(L/2) + Wload(L)/ Sin theta and the result was a smaller value. I'm really confused as to when to use length and when not to.
Hello, I wanted to know why we don't consider any y forces, X forces, or torque from the supporting cube touching the ground. It seems to make sense to me that the whole structure would topple over if there are no forces? Thanks
There are x and y forces where the post in embedded in the ground, but those forces would not add torque, since the distance from the point of rotation are considered "zero".
I love all your videos, but can you tell me why "d3" was really necessary? Isn't everything just acting at the full radial length or at half the radial length? I understand the need for the angle (sin 30), but couldn't all the torques just be set up quicker/easier as +mgr +MGr/2 and -Trsinϴ from the FBD without d3 since the cable provides torque at the full radial distance anyway?
Can the tension on the cable be resolved in the vertical direction then use the resolved force in calculating the torque about the same point you chose? Thank you.
If you are asking if we can use the components of the tension to find the torque, the answer is yes. Give it a try and see if you get the same answer (best way to check).
First of all: thank you. But I have to point out Two important things. The teacher drew (min. 0:50) the pivot...well. But he explained us that putting this pivot he eliminates the two unknown forces (reactions X and Y). Sorry it only eliminates another torque. The reactions stay or remain. Please just note that this force doesn't (or pair of forces Rx and Ry don't) produce moment or torque because its line of action passes through the fulcrum or pivot axis. . . Sorry about my English.
@@gokublack9080 if i remember four years ago, it went decent lol. this material was on the test and i was able to get it right, but i suck at physics. thank god i dont have to use it for my job
I solved this problem using the vertical projection of T multipied by d1 as the torque as the horizontal component of T is pointing along the beam and seems to not contribute to the torque of the beam and got a different answer? Could you please clear up how the tension force is acting at distance of d3? How would this make sense in terms of vectors?
Unless we consider these as vectors, the direction is not important if we remain consistent. However due to the possible confusion, I tend to hold to the vector definition of positive and negative in more current videos.
Absolutely amazing! I've reached your videos and I love how you teach. Thanks! One question: The 'y' component of T does not make any torque?? You just take T?
You can solve it either way. 1) Take the y component of T but then the distance d3 becomes the whole length L, or 2) take T and d3 becomes the distance as shown in the video.
I always thought that defining positive rotation to be clockwise seemed more logical and intuitive - after all, that's the "natural" rotation since the sun rotates that way, and we tend to stick to positive values as much as possible since most people prefer positive values.
+Laurelindo It only matters when when computing vector products, then the direction of the torque matters and should be taken to be positive in the counter-clockwise direction. Otherwise it doesn't matter as long as you stay consistent.
Wow man, I got an Exam and this is included, and I don't know how to solve it. You cleared it up big time for me. Thank you! Small Q: If I converted the signs and made the clockwise -ve and the counter +ve, the answer will be the same; is it okay?
is there no Force from the wall? to make segma Fx=0, we have the component of the Tension force to the left side, and we must have an opposite one, so is it N? i know it's not important for torque since it passes through the pivot but i would like to know
Yes, there are additional force we are not considering in this problem (because they do not affect the torque). There are some examples in the playlist that show how they are calculated.
