A 20 kg uniform ladder 2.5 meters long rests against a smooth wall at an angle of 53 degrees with the horizontal. Calculate the Forces on the ladder exerted by the floor and the wall.
Thank YOU!!.. this video was PERFECT from Beginning to End!!.. I think it's safe to say that even a 5th grader could have learned from this.. :) .. You are very Clear and stepwise methodical.. and that's what WORKS.. :) .. thanks again...
Thank you so much! this was so helpful. I was thinking that we were going to resolving the triangle into its x and y components. you got yourself a new subscriber. keep up the good work man!
I got confused with the FN. Why you did not consider the force acting on x-Axis at that point since the ladder stands inclined to the ground. (other than Fs). That should be FNy and FNx, right?
We cannot really say that there is a 'horizontal' gravitational force, although I think I know what you might be getting at. The gravitational vector is just that, and it points toward the ground as drawn. It does not have components on both axes, since it is not slanted with the coordinate system established as it is (cos90 = 0). Gravity does play a part in the horizontal axis however, because it is embedded in the definition of friction (indirectly, as part of the normal force).
The coordinate system you used here is weird for me, why do you take the cos component for the gravitational force of the ladder instead of the sin component, and the sin component for the normal force? To me, coordinate system you set here, implies that you take the sin component of gravity (since it acts downwards) and the cos component of the normal force (since it acts to the right)
Because if u were to draw x and y component of Fg u would see that the force perpendicular to the ladder can be represented by Fgcosø (not actually theta symbol)
I use a Wacom Intuos tablet. I think most of the others on CZcams who make tutorial videos use similar equipment: www.wacom.com/en-us/products/pen-tablets/wacom-intuos
why don't you just use 2.5sin(53) to find the perpendicular radius to the force of the wall, and then divide 2.5 by 2 and do 1.25cos(53) to find the perpindulcar radius to the force of the ladder. would that be incorrect? it seems much easier to me
How the heck is the y component perpendicular to the ladder, it isn’t lol…this makes no sense and how is the x component of the fg force perpendicular to the ladder cause it isn’t either …like what
Hi, I’m sorry you’re having trouble with this explanation. If you’ll look carefully, you’ll see that none of the vectors I drew are perpendicular to the ladder. Our vectors are oblique to the ladder, since it is slanted and the coordinate axis that is easiest to work with is not. For example, the force of gravity is simply straight down as it always must be.
Thank YOU!!.. this video was PERFECT from Beginning to End!!.. I think it's safe to say that even a 5th grader could have learned from this.. :) .. You are very Clear and stepwise methodical.. and that's what WORKS.. :) .. thanks again...
Thank you so much! this was so helpful. I was thinking that we were going to resolving the triangle into its x and y components. you got yourself a new subscriber. keep up the good work man!
Thank you very much for this excellent explanation.
You are amazing👌
Thank you very much 🙏
Thanks man!
how did you calculate the Fw? at the last part
I got confused with the FN. Why you did not consider the force acting on x-Axis at that point since the ladder stands inclined to the ground. (other than Fs). That should be FNy and FNx, right?
you're a saint
What if the wall also has friction
There would be an additional force in posetive y axis
Helpful tnx sir
why is the horizontal gravitational force not a part of the horizontal forces?
We cannot really say that there is a 'horizontal' gravitational force, although I think I know what you might be getting at. The gravitational vector is just that, and it points toward the ground as drawn. It does not have components on both axes, since it is not slanted with the coordinate system established as it is (cos90 = 0). Gravity does play a part in the horizontal axis however, because it is embedded in the definition of friction (indirectly, as part of the normal force).
The coordinate system you used here is weird for me, why do you take the cos component for the gravitational force of the ladder instead of the sin component, and the sin component for the normal force? To me, coordinate system you set here, implies that you take the sin component of gravity (since it acts downwards) and the cos component of the normal force (since it acts to the right)
Because if u were to draw x and y component of Fg u would see that the force perpendicular to the ladder can be represented by Fgcosø (not actually theta symbol)
very helpful, thank you
Nob
why did u use fg and fw as the forces in the torque equation
Those are the only forces that are a nonzero distance away from the 'hinge' that I chose, namely the point where the ladder meets the floor.
@@StephanPichardo thx
Which software do u use
It's Microsoft OneNote along with quicktime player on Mac for recording,
@@StephanPichardo thanks but are they free
@@asdfghjkl6506 Quicktime comes with MacOS and I believe OneNote is a free download from the Mac and Windows App Store.
@@StephanPichardo thanks
how do you draw so accurately
I use a Wacom Intuos tablet. I think most of the others on CZcams who make tutorial videos use similar equipment: www.wacom.com/en-us/products/pen-tablets/wacom-intuos
why don't you just use 2.5sin(53) to find the perpendicular radius to the force of the wall, and then divide 2.5 by 2 and do 1.25cos(53) to find the perpindulcar radius to the force of the ladder. would that be incorrect? it seems much easier to me
How the heck is the y component perpendicular to the ladder, it isn’t lol…this makes no sense and how is the x component of the fg force perpendicular to the ladder cause it isn’t either …like what
Hi, I’m sorry you’re having trouble with this explanation. If you’ll look carefully, you’ll see that none of the vectors I drew are perpendicular to the ladder. Our vectors are oblique to the ladder, since it is slanted and the coordinate axis that is easiest to work with is not. For example, the force of gravity is simply straight down as it always must be.