People commenting on this are absolutely correct, in the final question T should be multiplied by cos24 not divided like I have done. Moment of T : T x 2.5 x cos24 (2.5cos24 is the perpendicular distance to the pivot) Moment of F = F x 8 = 8F Equating these then allows you to solve for F since you already know T.
@@fohasajid2299 My method went wrong because of the step where I calculated F2 which just doesn't make logical sense. F2 is actually T * cos24 not what I wrote and that is why it went wrong (F2 is a component of T, T isn't a component of F2)
Just remember everyone is different. Some people find the moments concents very intuative and so need more challenging material, some find the concepts very challenging and it all looks like gibberish! I'm trying to have some materials for all of the above!
I sometimes struggle on the harder questions like with laptops being at angles opening. I find it hard to recognise some of the angles and perpendicular line of action. The ones that arent apparent. I dont know why, I just hope it clicks like the rest did at some point. Videos like these really help. Thank you.
it's been about 5 years since then so can you do this video again? with newer hard questions introduced and also do the last one again since u made a msitake in this video
Q2 is definetely wrong, it is tcos24 and not the other way round. The other way of solving it is by using the perpendicular distance for force t to pivot to get moment, just like the question before that
for the last question, can you not find that the horizontal component of T is T*tan24, therefore you can use pythag (root T² + T*tan24²) to find the hypotenuse which is the perpendicular distance from the pivot
2L is already perpendicular to the force at the end so there is no need to resolve it. L is not perpendicular to the weight so you have to find the component of L that is perpendicular to weight.
Thank you for your video. I was wondering, why is there a reaction force at the pivot, O? I'm confused since the force V acts in an anticlockwise direction and the force of gravity G acts in a clockwise direction, so why would we need a reaction force to balance them out? And why is the reaction force at the pivot point an upward force? Is that always the case at a pivot point? Thank you in advance.
So addressing those backwards. The reaction force at the pivot is not always upwards, it just happens to be in this case. In reality the pivot force is in whatever direction required to keep the object in equilibrium. In this case because all the forces are up or down, in this case it ended up being upwards. Also remember it is not just the moments that need to be in equilibrium, the forces also need to balance out. If you look at the magnitudes of the other forces you can see they don't balance without a reaction force.
I wouldn't say so. Tension arises when an object had been stretched and acts to unstretch the object. I think it is very unlikely the car bonnet is being significantly stretched!
That is what the highest difficulty questions in the exams always do. Show you something in a context you haven't seen before to see how well you really understand what you are doing!
7:14 how are you using f in the calculation its not resolved into vertical and horizontal forces this just looiks like a copy paste from markscheme to me
Wait so since your last ans is wrong, we can't assume that an opposite force will be acting on the left end too? the one you called F2? since according to the Ms, T is the force itself and not a component
Yeah that is where I went wrong introducing F2. I should have just found the distance perpendicular to T from T to the pivot and multiplied T by that distance which as you say is the actual force not a component.
Reaction forces don't act at the centre of mass, they act at the point of contact between two objects. (The only reason you sometimes see reaction forces acting at the centre of mass is that the centre of mass happens to be above the mid-point of the contact surface)
It tells you in the question (that part is cut off the screen at 8:52, you can see that earlier in the video. You can call the length anything you like, the key is that the centre of mass is in the middle so if the length is 2L, that means the centre of mass is L from the end (L will be the distance you multiply weight by to find its moment about the pivot)
Your last question is wrong. You don't need to make another triangle at the top, calculate the component of perpendicular to the pivot, your component isn't perpendicular so it cant be a moment.
I accept it’s wrong I’ve seen also on the mark scheme that it’s wrong but I also get what he gets when he puts 14000 over cos 24 rather than times it can you explain why it’s cos 24 x 14000 instead
@@fohasajid2299 yes. There is no F2. The concept of F2 is wrong. The only force there is tension and the component of it in perpendicular to the rectangle thingy is the only one there is.
People commenting on this are absolutely correct, in the final question T should be multiplied by cos24 not divided like I have done.
Moment of T : T x 2.5 x cos24 (2.5cos24 is the perpendicular distance to the pivot)
Moment of F = F x 8 = 8F
Equating these then allows you to solve for F since you already know T.
Pin this comment to the top, its important!
your method looks perfectly fine too then why is it wrong
@@fohasajid2299 My method went wrong because of the step where I calculated F2 which just doesn't make logical sense.
F2 is actually T * cos24 not what I wrote and that is why it went wrong (F2 is a component of T, T isn't a component of F2)
Harder questions ? Why do you mean by that ? Every question in this chapter is hard ?
Just remember everyone is different. Some people find the moments concents very intuative and so need more challenging material, some find the concepts very challenging and it all looks like gibberish! I'm trying to have some materials for all of the above!
I sometimes struggle on the harder questions like with laptops being at angles opening. I find it hard to recognise some of the angles and perpendicular line of action. The ones that arent apparent. I dont know why, I just hope it clicks like the rest did at some point. Videos like these really help. Thank you.
it's been about 5 years since then so can you do this video again? with newer hard questions introduced and also do the last one again since u made a msitake in this video
I might get to that at some point but there are quite a lot of ideas on my to do list before that!
Q2 is definetely wrong, it is tcos24 and not the other way round. The other way of solving it is by using the perpendicular distance for force t to pivot to get moment, just like the question before that
thanks
why not his method?
thought i was going crazy 💀
for the last question, can you not find that the horizontal component of T is T*tan24, therefore you can use pythag (root T² + T*tan24²) to find the hypotenuse which is the perpendicular distance from the pivot
then do the hypotenuse*2.5 =8F and then solve for F
Loved this, please do lots more
7:07 why do u resolve the distance of L but not two 2L?
2L is already perpendicular to the force at the end so there is no need to resolve it. L is not perpendicular to the weight so you have to find the component of L that is perpendicular to weight.
@@burrowsphysics ofc, thanks for replying
Thank you for your video. I was wondering, why is there a reaction force at the pivot, O? I'm confused since the force V acts in an anticlockwise direction and the force of gravity G acts in a clockwise direction, so why would we need a reaction force to balance them out? And why is the reaction force at the pivot point an upward force? Is that always the case at a pivot point?
Thank you in advance.
So addressing those backwards. The reaction force at the pivot is not always upwards, it just happens to be in this case. In reality the pivot force is in whatever direction required to keep the object in equilibrium. In this case because all the forces are up or down, in this case it ended up being upwards.
Also remember it is not just the moments that need to be in equilibrium, the forces also need to balance out. If you look at the magnitudes of the other forces you can see they don't balance without a reaction force.
Thank you so much for your reply, it was very helpful! 😊
I know you posted this years ago but for the first part of the first question could you say tension as another force acting on the bonnet?
I wouldn't say so. Tension arises when an object had been stretched and acts to unstretch the object. I think it is very unlikely the car bonnet is being significantly stretched!
4:58 thank me later
thank you
Can you explain how you do this
i was so confident then i got hit with the archimedes one lol
That is what the highest difficulty questions in the exams always do. Show you something in a context you haven't seen before to see how well you really understand what you are doing!
I did this question just now and my test is an hour. Man this stuff is hard
Mainly just intimidation tbh… it’s much easier than the other two questions in my opinion
Uh 1:50 How'd you get that to be 0.35m
says it in question if u look before
@@TheSweetZoltMan Actually found the same question online, but thanks anyway, I just slipped through the video :/
Saved my life thank you
7:14 how are you using f in the calculation its not resolved into vertical and horizontal forces this just looiks like a copy paste from markscheme to me
Why would you resolve it vertically and horizontally if it is already perpendicular to the distance to the pivot?
@@burrowsphysics exactly
Can i get the pdf of these questions?
Wait so since your last ans is wrong, we can't assume that an opposite force will be acting on the left end too? the one you called F2? since according to the Ms, T is the force itself and not a component
Yeah that is where I went wrong introducing F2. I should have just found the distance perpendicular to T from T to the pivot and multiplied T by that distance which as you say is the actual force not a component.
in the second question, F=19.6
F= mglcos60/2lcos60 = 4*9.8 / 2 = 19.6
No, because F is perpendicular to the pivot. The torque due to F is just FL not FLcos60
Great content
Where can I find these types of questions?
On the first one why isn't there a reaction force where the weight acts from?
Reaction forces don't act at the centre of mass, they act at the point of contact between two objects. (The only reason you sometimes see reaction forces acting at the centre of mass is that the centre of mass happens to be above the mid-point of the contact surface)
@@burrowsphysics thank you
Would be nice if we could see the assessment paper
How did you assume in the first question that the system is in equilibrium??
Nvm that was a dumb question...
@@SabrinaXe ahahaa
8:52 Sir, why is it 60?
Also, how can you assume the length is 2L?
It tells you in the question (that part is cut off the screen at 8:52, you can see that earlier in the video.
You can call the length anything you like, the key is that the centre of mass is in the middle so if the length is 2L, that means the centre of mass is L from the end (L will be the distance you multiply weight by to find its moment about the pivot)
For a uniform body the centre of mass is in the middle of the object
Your last question is wrong. You don't need to make another triangle at the top, calculate the component of perpendicular to the pivot, your component isn't perpendicular so it cant be a moment.
It is perpendicular and you are incorrect
Amazing videos, thank you
I also get 14000 divided by cos 24 but in the mark scheme it’s 14000 x cos 24 and I don’t understand why could you explain please
See my comment on the video (My answer is wrong and the mark scheme is correct!)
too eazy. U should try doing harder questions next time
at 9:11, the vertical F needs to be Fsin60 NOT Fcos60 as it is turning away from the angle you have drawn.
No, Fcos60 is correct as the vertical component of force F is the adjacent side
Further explain
Can you clarify which part? Thanks!
I'm 12
we don't care
@@alishashuikh7632 yup 13 now
@@mattyABC1 we dont care
@@bombi1591 14 now
@@mattyABC1 We care. We care so much. Keep it up!
last question is wrong
I accept it’s wrong I’ve seen also on the mark scheme that it’s wrong but I also get what he gets when he puts 14000 over cos 24 rather than times it can you explain why it’s cos 24 x 14000 instead
@@georgebanc6637 because its must be a fraction of the actual force how can a component have a higher value ?
@@Ogiiishred so T is the only force acting and the way he made it sound like T was s component of a force which was opposite to F is wrong?
@@fohasajid2299 yes. There is no F2. The concept of F2 is wrong. The only force there is tension and the component of it in perpendicular to the rectangle thingy is the only one there is.
@@harinachiappansubramanian9206 i got it correct with f2, just the fact that Tcos(24) = f2 and i did not divide and got 4060N