we need more explanation about how you came up with the assumptions made at the beginning of the video, FAx FAy FBx why they go that way, and why FB doesnt have a foce in y direction
the best way to understand why there is NO vertical force at B... is to imagine that the boom is NOT connected to the wall at B... the boom only Contacts the wall at B but is NOT connected to B... so, if it's NOT connected to the wall then only Point A is responsible for the VERTICAL force... So now you can rotate the boom up and down around Point A.... because Point A is like a Hinge... and when the boom is lowered to the point where it makes contact with the wall at B.. the only force is the wall pushing BACK on the boom in the X-direction... Hope that helps...
Mr. Michel, may I ask two questions: why is the force at the point A comprised of two components? And before actually seeing your video, I solved the problem myself, plugging the value of 9.81 for the acceleration due to gravity, which made my answers have tectonic discrepancy ( FnB = 760,275 N and FAy= 147,150 N.
The value of g depens on where on the Earth you are. It is 9.81 at the poles, but less everywhere else. (9.78 at the equator). Every force can be expressed as the sum of two components.
@MichelvanBiezen true but that change would be applicable to every component within the torque equation and could be factored out and essentially canceled out
i want to learn how to solve problems like this what is a good book for starters i work with cranes at work all the time it would be cool to learn advance stuff like this what do you recommend for starters ?
When torque is considered as a vector, CCW is always positive and CW is always negative. But if you don't consider torque as a vector it really doesn't matter (as long as you remain consistent throughout the problem.
You can choose any pivot you want, (it is arbitrary), however that said, picking a particular pivot point will eliminate any unknown forces who's line of action goes through the pivot point when calculating the moments (torques). Therfore some are better than others.
How can you just add the torques like that? The point of application is along the crane, and the lengths given are along the x-axis? If that's the case, wouldn't we need the angle with which the crane is pointed since the gravitational force acts at an angle and not just perpendicular to the crane (the 'arm')?
you have the distance from the line of action of the forces to the points where the line of action is perpendicular to the pivot point. Since we know this distance, we don't need to know the angle of the crane.
Professor: Do you have a similar video measuring the force necessary to pull up such a weight at a constant speed, and/or with little acceleration?? Thanks. K.S.
Two 1 kg masses hanging from both sides of a pulley, once set into motion (one moving up and the other moving down) will remain in motion at a constant speed (ignoring friction and air resistance).
@@MichelvanBiezen Professor, please see my new related comment below & respond, thanks. K.S.: (Physics Pulley Problems With Static Friction,) ruclips.net/video/30MV8gnWXEs/видео.html My comment:: at 8 assuming your weights are suspended on both sides of the pulley, then both sides need to be 20kg for the moving threshold, thus your friction coefficient is 1 (ONE); then if you add 2 kilos to one side for it to move, do you calculate the acceleration as follows:: Second example is for Apollo-Saturn:: (22)(9.8)-(1)(20)(9.8)==215.6-200==15.6 which divided by 42===0.37 m/s/s acceleration??? (3750)(9.8)-(1)(3000)(9.8)==36750-29400==7350 divide by 6750==1 m/s/s==a of Saturn-V Now, if you plug this 1m/s/s Apollo acceleration in a "Displacement Calculator", you'll see that in 10 seconds, Saturn-V could have been pulled up only 50 meters (half of its launch-tower-height), NOT the 100+ meters shown in the videos of its fastest Apollos, if it had had been hanging on one side of a [friction-less] pulley with a weight of 3,750 tons on the opposite side of the pulley. !!! Thus, as a rocket, it would NOT have moved at all, not even one inch!!! "Displacement Calculator" is linked under the description of: RUclips:: ("Solving-Apollo-Enigma-2")
Question: Sir, it is very counter-intuitive, at least to me, that Fbx causes a counter clockwise force. If it's toward the right, it should contribute a clockwise torque, same as those weight... But then the question itself would not make sense since the torque are no longer balanced. What do you think? Thank you for answering.
Awesome! Ideal for students oriented towards self studying.
Glad you think so!
Great & honest teaching. Thanks for sharing your expertise.
Our pleasure!
Simple, clear and concise, thank you.
we need more explanation about how you came up with the assumptions made at the beginning of the video, FAx FAy FBx
why they go that way, and why FB doesnt have a foce in y direction
but other than that it's great.
im very thankful for what you've done!
the best way to understand why there is NO vertical force at B... is to imagine that the boom is NOT connected to the wall at B... the boom only Contacts the wall at B but is NOT connected to B... so, if it's NOT connected to the wall then only Point A is responsible for the VERTICAL force... So now you can rotate the boom up and down around Point A.... because Point A is like a Hinge... and when the boom is lowered to the point where it makes contact with the wall at B.. the only force is the wall pushing BACK on the boom in the X-direction... Hope that helps...
Thanks for that explanation, Philip. I figured that must be what he was on about, but wasn't entirely sure of it.
Mr. Michel, may I ask two questions: why is the force at the point A comprised of two components? And before actually seeing your video, I solved the problem myself, plugging the value of 9.81 for the acceleration due to gravity, which made my answers have tectonic discrepancy ( FnB = 760,275 N and FAy= 147,150 N.
The value of g depens on where on the Earth you are. It is 9.81 at the poles, but less everywhere else. (9.78 at the equator). Every force can be expressed as the sum of two components.
Who else is here from ENED? Can we get an ENED bad comment chain?
yooooooo
Sir if we was to change the angle between the crane and horizontal would it affect anything?
Yes it would. The torque on the crane increases when the boom of the crane becomes horizontal.
@MichelvanBiezen true but that change would be applicable to every component within the torque equation and could be factored out and essentially canceled out
And sorry I meant if we change the angle between the bracket and the horizontal. Not the actual crane as if it were put on an inclined
Thanks love you
i want to learn how to solve problems like this what is a good book for starters i work with cranes at work all the time it would be cool to learn advance stuff like this what do you recommend for starters ?
There are many good physics books available. Any one would work. Have fun.
Does ccw is alway positive direction and cw is alway negative direction,teacher???
When torque is considered as a vector, CCW is always positive and CW is always negative. But if you don't consider torque as a vector it really doesn't matter (as long as you remain consistent throughout the problem.
Thank you teacher
What is the playlist name under which the crane related videos are there....pls reply..
You can find more videos like that here: MECHANICAL ENGINEERING 3 - EQUILIBRIUM OF RIGID BODIES
Sir,I'm new at mechanics.Will you answer me a question?
Sir,at 2:35 why we didn’t take Fa but we only took Fb?What different that tbe pivot made?
You can choose any pivot you want, (it is arbitrary), however that said, picking a particular pivot point will eliminate any unknown forces who's line of action goes through the pivot point when calculating the moments (torques). Therfore some are better than others.
How can you just add the torques like that? The point of application is along the crane, and the lengths given are along the x-axis? If that's the case, wouldn't we need the angle with which the crane is pointed since the gravitational force acts at an angle and not just perpendicular to the crane (the 'arm')?
you have the distance from the line of action of the forces to the points where the line of action is perpendicular to the pivot point. Since we know this distance, we don't need to know the angle of the crane.
good jop 👍👍
Professor: Do you have a similar video measuring the force necessary to pull up such a weight at a constant speed, and/or with little acceleration?? Thanks. K.S.
The force required to pull of an object at constant speed is equal to the weight of the object.
@@MichelvanBiezen So, then two 1kg weights on both sides of a pulley should be able to pull each other down??
Two 1 kg masses hanging from both sides of a pulley, once set into motion (one moving up and the other moving down) will remain in motion at a constant speed (ignoring friction and air resistance).
@@MichelvanBiezen Professor, please see my new related comment below & respond, thanks. K.S.:
(Physics Pulley Problems With Static Friction,) ruclips.net/video/30MV8gnWXEs/видео.html
My comment:: at 8 assuming your weights are suspended on both sides of the pulley, then both sides need to be 20kg for the moving threshold, thus your friction coefficient is 1 (ONE); then if you add 2 kilos to one side for it to move, do you calculate the acceleration as follows:: Second example is for Apollo-Saturn::
(22)(9.8)-(1)(20)(9.8)==215.6-200==15.6 which divided by 42===0.37 m/s/s acceleration???
(3750)(9.8)-(1)(3000)(9.8)==36750-29400==7350 divide by 6750==1 m/s/s==a of Saturn-V
Now, if you plug this 1m/s/s Apollo acceleration in a "Displacement Calculator", you'll see that in 10 seconds, Saturn-V could have been pulled up only 50 meters (half of its launch-tower-height), NOT the 100+ meters shown in the videos of its fastest Apollos, if it had had been hanging on one side of a [friction-less] pulley with a weight of 3,750 tons on the opposite side of the pulley. !!! Thus, as a rocket, it would NOT have moved at all, not even one inch!!! "Displacement Calculator" is linked under the description of: RUclips:: ("Solving-Apollo-Enigma-2")
Sir what if the weight of mg is not given?
mg needs to be a "given".
Question: Sir, it is very counter-intuitive, at least to me, that Fbx causes a counter clockwise force. If it's toward the right, it should contribute a clockwise torque, same as those weight... But then the question itself would not make sense since the torque are no longer balanced. What do you think? Thank you for answering.
Fbx causes a counter clockwise torque (not a counterclockwise force) relative to pivot point A.