Mr. Michel, I believe you should habe drawn an accurate and self-explanatory free body diagram. Morever, why did you even take into consideration the right side of the ladder only while the tension also exists from the left side?
please note something is wrong in this problem.... since when we consider only the right portion of the ladder and sum forces in the X-direction they won't add up to zero which means that the ladder is moving along X-direction in the horizontal plane..... i guess the solution is that the reactions at A and B must be along the direction of the members AC and BC respectively and thus we will obtain equilibrium along X and the answer for T will change it would be 218.75kg*m/sec2 or simply 21.875 kg..... please note this..... since we're putting our content to the public we must provide correct solutions otherwise we'll teach some wrong concepts
@@MichelvanBiezen when we do the free body diagram to get the tension in the cable the only force in the X-direction is the tension T so SUM of FORCES ALONG X is not equal to zero and thus their is no equilibrium what means the ladder is moving under the action of the T force which is definitely not the case..... The reactions of the supports at A and B should be in the direction of the members this is the only way to obtain equilibrium
I encountered a similar problem to this. And it asks me to find the magnitude of the force exerted of the right side of the ladder on the left side. Is it the same with the tension acquired in this problem?
Hi Prof., may I ask @minute 8:01, won't we need to take the sin(75.52°) of the upper triangle's 2M (hypotenuse) to get its vertical distance that is then perpendicular to tension "T"? Nevertheless your video are extremely helpful, some of your examples I originally don't even know where to start. Thank you
My apologies. @ Minute 6:05 you indeed already specified d3 to be the vertical component of upper triangle's hypotenuse. The section in question should be @ minute 8:10, F(b) was only "(3/8)mg" and not "(3/8)mg(sin75.52)" the vertical component of F(b) which then multiple by "1" the distance from point C to the line of action of F(b).
These calculations of torque performed in plane XOY or in two dimensions. Is the calculation need to be performed in third dimension, plane ZOY, in order to can save someone who use it?
You only need to calculate the torque on one side of the ladder in order to find the tension. There are only two forces that act on the right side of the ladder, T and Fb
@michel van biezen you said the tension is an internal force so was excluded when taking moments about pivot A but then why did you take it into account when taking moments about pivot C?
That is correct. When calculating the moment about A, we only considered the external forces, on the ladder-person system. When we calculated the moment about C (which is a movable joint) we considered all the forces that can make the member to the joint move.
@@MichelvanBiezen sorry, but i did not get it. What do you mean by "C" is a movable joint and how does that make an internal force to be taken into account? please help me understand
Can you please give us some more example of tension force like this one as when we take the right side for point c, that part is quite confusing for some of us, that would very helpful to understand how it works, thanks
I assume that if we used the LEFT side of the ladder to find the TENSION, "T", then we would arrive at the same value...... I'll have to try that.. :) ....
@@MichelvanBiezen hello I was wondering what this comment meant? why would T be not an external force? im guessing it's because on pivot A, the tension from the right gets cancelled by the tension on the left, but on the top pivot, taking only one side, it cannot be cancelled since we are only looking at one side? or is it because the whole ladder is the pivoting object when using pivot A?
I'm doing a project for uni of this same type of problem, the same type of ladder, but I'm trying to incorporate the effects of friction at the feet and I'm really stuck, I can't find any information on how to deal with this problem. Do you have any idea where I can get more information on this? Any info I find is just about a simple ladder leaning against a wall.
If there is no friction between the ladder and the floor, and the ladder doesn't move when the person on the ladder is not moving, then there is no net force in the horizontal direction.
@@MichelvanBiezen I appreciate the reply, but part of my investigation is to look at the deflection of the legs of the ladder when used on a low friction floor such as vinyl laminate flooring. I think I'm going to have to split my investigation up and do an analysis on the full ladder assuming no friction to get an overall idea of what's happening, then consider friction just for the legs that deflect and look at them in isolation
Exactly. That is how you want to approach it. After you determine the vertical forces and the tension on the rope. Pick a rotation point at the connection of the rope and the ladder and calculate the torque, to see if there are any horizontal forces between the ladder and the floor.
Since he is standing 3/4 the distance up the ladder. His center of gravity will be 3/4 the distance (1 m) from the and to the middle of the ladder (in the horizontal direction).
The tension (magnitude) is the same on both side. However, the rope is pulling to the left with respect to the right side of the ladder, and the rope is pulling to the right with respect to the left side of the ladder.
Little confused.You say that because of T is vectoral,which side of the rope has positive/negative tension?And how can you take an only one side of forces to calculate the "Net torque equals zero" equation?(left sides net torque is not zero.)
Would the tension on the left side equal the tension on the right side? Intuitively I want to say it must but when I calculate the tension I find it to be 221N on the left
Hello sir I'm from Vietnam . Question: find the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half Can you help me thanks sir
This is my question. Man of 58kg is 2m up the ladder. The length of the ladder is 2.5. The legs are 1.8m apart. The answer for the tension is around 180m By using your method, my tension is 193, which is wrong.
when you find torque C;so why didnt you subtract Force A on that equation? I really hope you will answer my question🙏🙏🙏
Mr. Michel, I believe you should habe drawn an accurate and self-explanatory free body diagram. Morever, why did you even take into consideration the right side of the ladder only while the tension also exists from the left side?
There are many ways in which this problem can be solved. We showed one method.
please note something is wrong in this problem.... since when we consider only the right portion of the ladder and sum forces in the X-direction they won't add up to zero which means that the ladder is moving along X-direction in the horizontal plane..... i guess the solution is that the reactions at A and B must be along the direction of the members AC and BC respectively and thus we will obtain equilibrium along X and the answer for T will change it would be 218.75kg*m/sec2 or simply 21.875 kg..... please note this..... since we're putting our content to the public we must provide correct solutions otherwise we'll teach some wrong concepts
Video is correct. Note that the coefficient of friction = 0 and thus there cannot be any net force on the ladder in the x-direction.
@@MichelvanBiezen when we do the free body diagram to get the tension in the cable the only force in the X-direction is the tension T so SUM of FORCES ALONG X is not equal to zero and thus their is no equilibrium what means the ladder is moving under the action of the T force which is definitely not the case..... The reactions of the supports at A and B should be in the direction of the members this is the only way to obtain equilibrium
I encountered a similar problem to this. And it asks me to find the magnitude of the force exerted of the right side of the ladder on the left side. Is it the same with the tension acquired in this problem?
It would make more sense to find the tension at the connection point. We always need to reference any tension or force to where it is applied.
Does hinge force exist at the c point ? and Is it considered an internal force ?
Hi Prof., may I ask
@minute 8:01, won't we need to take the sin(75.52°) of the upper triangle's 2M (hypotenuse) to get its vertical distance that is then perpendicular to tension "T"?
Nevertheless your video are extremely helpful, some of your examples I originally don't even know where to start.
Thank you
Isn't that what we did in the video?
My apologies. @ Minute 6:05 you indeed already specified d3 to be the vertical component of upper triangle's hypotenuse.
The section in question should be @ minute 8:10, F(b) was only "(3/8)mg" and not "(3/8)mg(sin75.52)" the vertical component of F(b) which then multiple by "1" the distance from point C to the line of action of F(b).
These calculations of torque performed in plane XOY or in two dimensions. Is the calculation need to be performed in third dimension, plane ZOY, in order to can save someone who use it?
Can I ask why the tension is not incorporated when taking the torque with respect to (A)?
The tension of the rope is internal to the system and therefore does not produce a torque. (Only external forces can do that).
Hi why is the person not experiencing a normal force as he/she is standing on one of the steps of the ladder?
He/she is, but the normal force has no bearing on the forces we are looking for.
@@MichelvanBiezen Oh I see is it because the ladder has a negligible mass?
The mass of the ladder is "negligible" indeed.
when you find the torque at C, why did you not include F in a and the mass mg? thank you
You only need to calculate the torque on one side of the ladder in order to find the tension. There are only two forces that act on the right side of the ladder, T and Fb
The tension is the same regardless which side you calculate it on.
@michel van biezen you said the tension is an internal force so was excluded when taking moments about pivot A but then why did you take it into account when taking moments about pivot C?
That is correct. When calculating the moment about A, we only considered the external forces, on the ladder-person system. When we calculated the moment about C (which is a movable joint) we considered all the forces that can make the member to the joint move.
@@MichelvanBiezen sorry, but i did not get it. What do you mean by "C" is a movable joint and how does that make an internal force to be taken into account? please help me understand
thank you sir!!!! i understood all
Can you please give us some more example of tension force like this one as when we take the right side for point c, that part is quite confusing for some of us, that would very helpful to understand how it works, thanks
We do have a playlist on that topic: PHYSICS 17 TENSION AND WEIGHT
@@MichelvanBiezen Thanks a lot, your videos are really helpful
I assume that if we used the LEFT side of the ladder to find the TENSION, "T", then we would arrive at the same value...... I'll have to try that.. :) ....
Just did the left side and VOILA! it works.... 133 N ........ PHYSICS WORKS!
It is always satisfying when it does.
In calculating the torque at point A, why didn't you include the tension?
The tension T is an internal force and therefore does not cause a torque about point A. That requires and external force.
@@MichelvanBiezen Oh i remembered now, thank you!
@@MichelvanBiezen hello I was wondering what this comment meant? why would T be not an external force? im guessing it's because on pivot A, the tension from the right gets cancelled by the tension on the left, but on the top pivot, taking only one side, it cannot be cancelled since we are only looking at one side? or is it because the whole ladder is the pivoting object when using pivot A?
If we take torque about B then tension will create an anticlockwise torque then which other force on BC ladder will create clockwise torque
The tension will not form a torque about B
@@MichelvanBiezen why? because i thought you can choose any point as the axis of rotation
what if we friction coefficient is not equal to 0?
Then it becomes a more difficult problem as there are more forces involved (including the friction forces)
Can we calculate torque about point B
You can pick any point as the pivot point.
I'm doing a project for uni of this same type of problem, the same type of ladder, but I'm trying to incorporate the effects of friction at the feet and I'm really stuck, I can't find any information on how to deal with this problem. Do you have any idea where I can get more information on this? Any info I find is just about a simple ladder leaning against a wall.
If there is no friction between the ladder and the floor, and the ladder doesn't move when the person on the ladder is not moving, then there is no net force in the horizontal direction.
@@MichelvanBiezen I appreciate the reply, but part of my investigation is to look at the deflection of the legs of the ladder when used on a low friction floor such as vinyl laminate flooring. I think I'm going to have to split my investigation up and do an analysis on the full ladder assuming no friction to get an overall idea of what's happening, then consider friction just for the legs that deflect and look at them in isolation
Exactly. That is how you want to approach it. After you determine the vertical forces and the tension on the rope. Pick a rotation point at the connection of the rope and the ladder and calculate the torque, to see if there are any horizontal forces between the ladder and the floor.
Hi sir, at 3.00 you have said the distance is (3/4m). I am not very clear how you got that. Can u pls explain me? Thanks :)
Since he is standing 3/4 the distance up the ladder. His center of gravity will be 3/4 the distance (1 m) from the and to the middle of the ladder (in the horizontal direction).
So,does it mean tension at both sides of the rope are equal?
*And magnitudes
The tension (magnitude) is the same on both side. However, the rope is pulling to the left with respect to the right side of the ladder, and the rope is pulling to the right with respect to the left side of the ladder.
Little confused.You say that because of T is vectoral,which side of the rope has positive/negative tension?And how can you take an only one side of forces to calculate the "Net torque equals zero" equation?(left sides net torque is not zero.)
Draw a free body diagram at each end, use vectors for the forces and you'll see. (Let me know if you don't know what a free body diagram is).
I got torque but i've never drawn a free body diagram with forces that does not directly acting on object.
Calculating this problem is wrong ..No perpendicular on all forces were considered
Why do we ignore torque = rFsin(theta) and simply use t = rF ?
Since there is no friction on the floor, there is only a normal force between the ladder and the floor.
Would the tension on the left side equal the tension on the right side? Intuitively I want to say it must but when I calculate the tension I find it to be 221N on the left
Never mind I left out a force it now balances to 133N on each side
Hello sir I'm from Vietnam .
Question: find the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half
Can you help me thanks sir
Bảo Khang Nguyễn quen quen ta :)
i got 136 instead of 133
sorry, I realised that you need to take g is 9.81 instead of 10
this is a sham, d3 should be 2m instead of sin thetha. Used his method on my Mastering Physics and i got the wrong answer
The video is correct
This is my question. Man of 58kg is 2m up the ladder. The length of the ladder is 2.5. The legs are 1.8m apart. The answer for the tension is around 180m
By using your method, my tension is 193, which is wrong.
Where is the location of the rope holding the two sides of the ladder? (and is the mass of the ladder = 0)?