Thank you so much!! Physics is one of those things where it will not make sense until that moment where it all of the sudden clicks.. and to do that you need a good teacher. So thank you!
I'm in an online AP Physics class, and so there isn't a teacher for me to ask questions to about physics stuff. Tension was taking me forever to learn, but THANK GOD for this video to teach me tension problems! I get it now, thanks a lot Dr. Holt :D
Dr. Holt, When you solved for FAB you divided both sides of the equation by the wrong value. You should divided both sides by -.8 to get the value that you obtained because if you divide by -.857 you will get FAC=.933 FAB.
Your approach is correct. However, if we don't round off to get the approximate values for theta1 & theta2, we can get a better result. tan(theta 1) = 3/4 => sin(theta1) = 3/5, cos(theta1) = 4/5 . tan (theta 2) = 3/5 => cos (theta2) = 5/sqrt(34) , sin(theta2) = 3/sqrt(34). The system of equations will be : resolve (vertically) : 3/5* Fab + 3/sqrt(34)* Fac = 30g . resolve (horizontally) - 4/5 * Fab + 5/sqrt(34)* Fac = 0 => Fab = 250/ 9 * g = 272.2 (N) , Fac = 40*sqrt(34) * g/ 9 = 253.97 N
I think the answers for Fab and Fac given in the video are incorrect, since their vector sum is not even approximately equal to Fad, which is 294 N. For Fab and Fac, I get 163 N and 191 N, which on vector addition appropriately give SQRT(163^2+191^2+2(163)(191)cos(37 deg +31 deg)) = 294 N.
So on a 500' 65° decline..600plb weight..... Tree to tree guy wire. Using 5/8 braided cable. The tree at the top, the wire is attached 15' up. On the bottom tree, the wire is attached 20' up from the ground. The 600plb object is 8' long and 5+1/2' wide and 3' tall in height. Will be suspended by ropes and three equally spaced tandem pulley trolleys. Where as the top of the 600plb object leading to the bottom of the guy wire will be 9'. How do I know how tight the guy wire needs to be so that the object doesn't hit the ground but ends within 20' of the bottom tree and about 1+1/2' from the ground? Who is the math god?
Newton's 3rd law is that for every action there is an equal and opposite reaction. The object isn't moving, so there are equal forces acting in opposite directions. If you imagine that the weight is on the y-axis of a graph, the AB vector is going in the negative x direction while the AC vector is going in the positive x direction.
No, he is actually correct in saying that "Fab = 1.07Fac" but was in error in cancelling out -0.875 in order to isolate Fab when he shouldve divided all by 0.8.
Two easy formulas that answer this type of problem every time without all of the equations. Each uses three angles. A,B and C. In the problem above the angles are: A = 30.96, B = 36.87, C = 112.17(angle C is the angle between A & B) Angle C is calculated as 180-(A+B) = 112.17 Also, FAD in this problem is 294. FAC =(FAD*Cos(B))/Sin(C) and FAB =(FAD*Cos(A))/Sin(C) Note: The answers will differ slightly than in the video because this method doesn't do rounding early in the problem(except for the angle measures) as is done in the video. You can eliminate rounding angles if you have a calculator that will store to variables then use the stored values in the calculation. Even if you must use the equations as in the video, these formulas are good for self-checking your answer.
Siphelele Kondlo He doesn't explain it very well at that point, he moves numbers over to the left side then they magically appear next to FAC? Pretty sure it looks wrong from hwere I'm standing considering he hasn't explained how that happened mathematically.
Joel Haire The thing is he moved all his numbers over to one side then writes FAB= 1.07FAC without even mentioning why. I know how to do mathematics but the explanation isn't very good. Hence why I said he doesn't explain it very well. you know in 5th grade mathematics and through most of school your told to show your working out, which would be ESPECIALLY important when teaching someone
Ghost572 I noticed as I watched the video carefully, it should state that Fac=1.07Fab using basic linear work, he's made a simple mistake and just written it the other way around, and also since he is teaching to lets say everyone that comes across this video on youtube that doesn't understand his concepts, should consider a minute or 2 at ending of this video for an explanation to the mathematics side of things so all people can comprehend.
I think there's a mistake. I got the same answer for F(ab) but my F(ac) is greater than F(ab). How could .8/.857=1.07???? I think your problem is right there...
He went the other way and solved for FAB by divided both sides by -.80. so the -.80 cancelled out leaving FAB = 1.07FAC. The math is right but he messed up his demonstration because he went one way of solving it in the demonstration but forgot that the answers to each step that he wrote down were from solving in the other direction. Do you understand what I mean?
Why do you round to a different number of decimal places in each number? Is there a recommended number of decimal places to use in order to get a more accurate answer? Should it not be consistent? Would it not be better to use all decimal places in the steps and only round them up in the final answer?
Why would you say if your calculator was set to radians it wouldn't work? It would most certainly work in radians, even if you mistakenly thought you were using degrees the whole time.
Here are my answers (I used t1-Fab t2-Fac and t3-Fad) For t1= 272.25 N T2= 253.98N T3 = 294 N In your final answer, when i check it (all forces on the x-axis should be equal to 0), yours is laying on 1... something. Mine has a closer value to zero.
The steps are perfect. However why would you replace with the numbers at such an early stage ? You are augmenting the marge of mistakes. It's better to handle that as the last step of the problem solving.
I used my procedure and your procedure and came up with the same answers which are: 610.74 N for the Tsub1 and 570.25 N for the second tension force. Only 294 N as an answer for the third tension force stood the same on both parts. How come?
A ladder is placed against a perfectly smooth wall, while the base rest on rough ground making an angle of 60 with the horizontal. A work man of mass 95 kg stands halfway up the ladder. Assuming the ladder does not slip on the floor, find: (i) The reaction between the wall and the ladder (ii) The direction of the reaction of the floor on the ladder
Hello Dr Holt, just want to point out just as @Noreen McGee- below and others that (-.8/-0.857)= 0.9335F(ac). Hope you can edit to reflect correction. Thanks for the video.
Thank you so much!! Physics is one of those things where it will not make sense until that moment where it all of the sudden clicks.. and to do that you need a good teacher. So thank you!
My test is on a few hours but now I know I'm going to get a good grade. Thank you sir.
I'm in an online AP Physics class, and so there isn't a teacher for me to ask questions to about physics stuff. Tension was taking me forever to learn, but THANK GOD for this video to teach me tension problems! I get it now, thanks a lot Dr. Holt :D
Dr. Holt,
When you solved for FAB you divided both sides of the equation by the wrong value. You should divided both sides by -.8 to get the value that you obtained because if you divide by -.857 you will get FAC=.933 FAB.
I agree with you Noreen McGee. Thanks for it because i was also confused about it. Thnx alot
I noticed that as well.
oh true
Zach Dissen I think he just accidentally wrote FAB instead of FAC
which is the real answer?
hi. I wanna say thank you. this still helps me in 2019.
You lost me at 3:30
This is not complicated at all. He is just being thorough with all his steps. This is the correct way to do these tension problems.
i have a tension problems; i just can't concentrate!
Thanks heaps, this was surprisingly hard to find an example where B and C are at the same height very helpful :) .
dude your a genius I was trying to find out how to do this for 2 friggin hours and the video was here the whole time ;D
how would you go about answering this when given 3 cables as opposed to 2
Your approach is correct.
However, if we don't round off to get the approximate values for theta1 & theta2, we can get a better result.
tan(theta 1) = 3/4 => sin(theta1) = 3/5, cos(theta1) = 4/5 .
tan (theta 2) = 3/5 => cos (theta2) = 5/sqrt(34) , sin(theta2) = 3/sqrt(34).
The system of equations will be :
resolve (vertically) :
3/5* Fab + 3/sqrt(34)* Fac = 30g .
resolve (horizontally)
- 4/5 * Fab + 5/sqrt(34)* Fac = 0
=> Fab = 250/ 9 * g = 272.2 (N) , Fac = 40*sqrt(34) * g/ 9 = 253.97 N
I think the answers for Fab and Fac given in the video are incorrect, since their vector sum is not even approximately equal to Fad, which is 294 N. For Fab and Fac, I get 163 N and 191 N, which on vector addition appropriately give SQRT(163^2+191^2+2(163)(191)cos(37 deg +31 deg)) = 294 N.
This video save my time and life!!!! thanks dude!!! i was stuck on this type of problem and was trying to figure it out for hours!!!
if I did not go to school or had not learn this already I would have had no idea what was going on...more detailed explanations would be lovely
Dr. Holt? the same one whos name is listed on my textbooks!??!?!?!?!?!
Your pic goes so well with that comment xD
GoPostal XDDDDDDDDDDDDDDDDD
So on a 500' 65° decline..600plb weight..... Tree to tree guy wire. Using 5/8 braided cable. The tree at the top, the wire is attached 15' up. On the bottom tree, the wire is attached 20' up from the ground.
The 600plb object is 8' long and 5+1/2' wide and 3' tall in height. Will be suspended by ropes and three equally spaced tandem pulley trolleys. Where as the top of the 600plb object leading to the bottom of the guy wire will be 9'. How do I know how tight the guy wire needs to be so that the object doesn't hit the ground but ends within 20' of the bottom tree and about 1+1/2' from the ground?
Who is the math god?
Newton's 3rd law is that for every action there is an equal and opposite reaction. The object isn't moving, so there are equal forces acting in opposite directions. If you imagine that the weight is on the y-axis of a graph, the AB vector is going in the negative x direction while the AC vector is going in the positive x direction.
It would be much easier to understand and match together if you didn't bother converting cos and sin. Just save it for the end next time
Lol, sitting with exact the same problem except for that I am given the angles. Thanks for the help!
No, he is actually correct in saying that "Fab = 1.07Fac" but was in error in cancelling out -0.875 in order to isolate Fab when he shouldve divided all by 0.8.
Two easy formulas that answer this type of problem every time without all of the equations. Each uses three angles. A,B and C. In the problem above the angles are: A = 30.96, B = 36.87, C = 112.17(angle C is the angle between A & B) Angle C is calculated as 180-(A+B) = 112.17 Also, FAD in this problem is 294.
FAC =(FAD*Cos(B))/Sin(C) and FAB =(FAD*Cos(A))/Sin(C) Note: The answers will differ slightly than in the video because this method doesn't do rounding early in the problem(except for the angle measures) as is done in the video. You can eliminate rounding angles if you have a calculator that will store to variables then use the stored values in the calculation.
Even if you must use the equations as in the video, these formulas are good for self-checking your answer.
at 6:49, i think he made a mistake, Fab is not = 1.07Fac, it should be Fac=1.07Fab. am i right?
I think you are very correct
Siphelele Kondlo He doesn't explain it very well at that point, he moves numbers over to the left side then they magically appear next to FAC? Pretty sure it looks wrong from hwere I'm standing considering he hasn't explained how that happened mathematically.
Ghost572 All physics is linear work, which everyone should have developed in year 11 or 5th form Maths. Cheers.
Joel Haire The thing is he moved all his numbers over to one side then writes FAB= 1.07FAC without even mentioning why. I know how to do mathematics but the explanation isn't very good. Hence why I said he doesn't explain it very well.
you know in 5th grade mathematics and through most of school your told to show your working out, which would be ESPECIALLY important when teaching someone
Ghost572 I noticed as I watched the video carefully, it should state that Fac=1.07Fab using basic linear work, he's made a simple mistake and just written it the other way around, and also since he is teaching to lets say everyone that comes across this video on youtube that doesn't understand his concepts, should consider a minute or 2 at ending of this video for an explanation to the mathematics side of things so all people can comprehend.
I realized later there was an error and if I get time, I may correct it. Sorry about that; however, I hope the approach that I use is helpful.
I think there's a mistake. I got the same answer for F(ab) but my F(ac) is greater than F(ab). How could .8/.857=1.07???? I think your problem is right there...
That's what I was saying too!
YES !!
He went the other way and solved for FAB by divided both sides by -.80. so the -.80 cancelled out leaving FAB = 1.07FAC. The math is right but he messed up his demonstration because he went one way of solving it in the demonstration but forgot that the answers to each step that he wrote down were from solving in the other direction. Do you understand what I mean?
Thank you so much Dr. Holt! Really helpful!
You are amazing! Life saver!!
ur the reason i passed m8 TY
Why do you round to a different number of decimal places in each number? Is there a recommended number of decimal places to use in order to get a more accurate answer? Should it not be consistent? Would it not be better to use all decimal places in the steps and only round them up in the final answer?
great job man, thanks alot.....doing some calculations for conductors on posts.....
R.I.P. headphone users :(
Yeah I see it too, easy enough to work around though & still a great video on thw whole. Thanks brent
great video. better than my Prof.
your equation for the EFx = F(AC) = 1.07(AB)
Glad im not the only one who noticed his mistake
InfamousColinGH Made my head hurt but I see what happened now, simple mistake.
Darshen Patel you are going right
Why would you say if your calculator was set to radians it wouldn't work? It would most certainly work in radians, even if you mistakenly thought you were using degrees the whole time.
this helped a lot:) do one on a inclined plane
can anyone explain to me when to use sin/cos in these equations,sorry but i forgot about this
when finding the x component of a force, you use cos. when finding a y component, you use sin
thanks for the reply, already got it tho.
A string is stretched by two equal and opposite forces 10N each. The tension in the string is.....?
This helped a lot, thank you.
Fac and fab values are inverted
Mind bending
+Aimee Sedlarik Seemed pretty damn obvious to me. Great video, helped me a lot
maybe he missed the the answer''',,, but the strategy is good and also the explanation method....
your really good in teaching,. keep it up :) thank you
Actually it is correct. 0.93FAB=1FAC. If you divide both sides by 0.93 you get FAB=1.07FAC or FAB=1.08FAC depending on how you round.
Here are my answers (I used t1-Fab t2-Fac and t3-Fad)
For t1= 272.25 N
T2= 253.98N
T3 = 294 N
In your final answer, when i check it (all forces on the x-axis should be equal to 0), yours is laying on 1... something. Mine has a closer value to zero.
Thank you so much. It's a great tutorial. :)
Love u
Thankyou so much!
thanks! very helpful.
The steps are perfect. However why would you replace with the numbers at such an early stage ? You are augmenting the marge of mistakes. It's better to handle that as the last step of the problem solving.
please. can you give a specific solution on how did get the
fAc=254 im so stock in rhat part..please..
At 5:54 there is an error in that -0.8/-0.857 = 0.93 not 1.07.
thank you this was really helpful
Dr.Holt, why didn't you inverse the tangent in theta(2) like you did while finding theta(1)?
+Matt Quinn how did he go from -294N+1.156 = 254.3N?
It's ok but i'll tell you that it's very very late
Thank you Doctor
Schrodie56, the Professor meant to divide both sides by 0.8 to get Fab= 1.07Fac.
I used my procedure and your procedure and came up with the same answers which are: 610.74 N for the Tsub1 and 570.25 N for the second tension force. Only 294 N as an answer for the third tension force stood the same on both parts.
How come?
I think its Fac=1.07Fab
Why the statics is a course in univercities, high school kids can solve them easily. Piece of cake!
Check your spelling first. You should go back to gradeschool
You don't need to go over all of this.... Just use this formula: Force(tension)=(Force applied)/(2sin(theta)) that's it...
+UzCulinary101 you just saved me so much time. thanks
EXACTLY! OK turns this mess off!
Hiii
Hi, so how would you calculate the tensions if the weight was bounced on that cable?
shouldnt it be -Fad and not -Fab. Since the tension for AD is -294???
Or am i seeing something wrong
Um. I'm pretty sure it's. Fac = 1.07Fab. Not the other way around. But still, great explanation.
thank you for the explanation :))
Hiii
thank you! very helpful
guys i think he is right they are many method to kill that i used my method but i got the same answer
A ladder is placed against a perfectly smooth wall, while the base rest on rough ground making an angle of 60 with the horizontal. A work man of mass 95 kg stands halfway up the ladder. Assuming the ladder does not slip on the floor, find:
(i) The reaction between the wall and the ladder
(ii) The direction of the reaction of the floor on the ladder
Cool.
why is there no tension at AD?
Confusing
Thanks Doc!
why is the x component AB negative?
Y Fx Fy and not Tx Ty. shud wire be tension ?asking for a help here
Hello Dr Holt, just want to point out just as @Noreen McGee- below and others that (-.8/-0.857)= 0.9335F(ac). Hope you can edit to reflect correction. Thanks for the video.
what if you only have tension one one rope and the angle it makes with the ceiling? the other two variables are unknown. weight is known.
myong2206 i thought I was the only one with that type of question
Wait so when you are trying to find out what Fac is equal to how do you know which side to solve for. Fac or Fab?
Did he switch fab and fac by accident when solving for fac?
doctur hult i miss yore smile
very useful video
thank you sir .
what happens if its being pulled down by A Lb and not kg
+alejandro cueva multiply LB to KG by 2.2 :) u should google it bro
why u made it complicate ?
@7:26 actually thats 1.114AC (you made addition mistake of 0.60+0.514=1.114)not 1.156 so that finally Tension in AC=263N
How to get Fab=1.07
kilonewtons?
where is the direction of tension
Tension is always pulling away from the point. If it went the other way it would be in compression.
He made a mistake at 5:54
Thank you
how come -294+1.156 is equals to 254.3
Arteinnian Grimes because it's 294/1.156=254.3..the variable was Fac which was multipled by 1.156 so to undo it you need to divide it out.
where does the -80fab come from ??
I'm confused
quesquil faut prendre pour baissair la tention jai 73.ans
it is nice
i like it.
the tension of AD is the weight :)
It's more easy to apply Lamy's law.
solving for Fab is 1.07.
He just solved for the wrong tension and wrote the other tension
it should have been .857/.8= 1.07
Wrong solution on 6:20. You should divide the forces by -.80 instead of -.857. Though however the answer is correct. :)
If the answer is correct, would the division also be correct?
use lami
2 equation 2 unknown
not an engineer, so how does the type of material considered in these scenarios?
Young's modulus
+James Elmore thanks man.