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# M1 help? Exercise 3B Question 9 watch

1. The question states that masses A and B are in a lift of mass 250kg. The lift is in constant acceleration upwards and the lift exerts a reaction force of magnitude 678N and 452N on masses A and B respectively. It also states the tension on lift is 3955N.
I looked at the solution bank and it is taking the reactions forces to be the weight of the masses and solves the problem by applying n2l on the lift only. I was taught that the reaction force is different to the weight of the object when in acceleration so the force exerted on the lift is equal to the weight of the masses of the 2 masses.
2. Hello ? Is anyone there to help ?😰
Hello ? Is anyone there to help ?😰
Yes but I've not got the book in front onf me. Can you type the whole question please? I'm not clear what you are asking.
4. Yeah here's a picture. It's question 9:
5. The solution is asumming that the reaction force is equal to the weight Albert (A) and Bella (B). The weight of A and B and the weight the weight of the life (250g) is the force acting downwards on the lift and not the reaction force exerted by the lift on A and B plus the weight of the lift.
6. (Original post by Muttley79)
Yes but I've not got the book in front onf me. Can you type the whole question please? I'm not clear what you are asking.
I posted the qs
I posted the qs
Thanks - so what is the bit you are unclear about? I think you are confusing the forces acting on the whole system to those 'internal' forces acting on the people.
Yeah here's a picture. It's question 9:
It really helps if you draw separate diagrams for lift questions. So draw 3 force diagrams for Albert, Bella and the system of the lift containing Albert and Bella.

Firstly, it's important to realise that all three of these objects will be moving with the same acceleration.

For Albert, the forces acting on him will be his weight and the force exerted on him by the lift. The forces acting on Bella will be similar.

For the lift containing Albert and Bella, you have the weights of the lift as well as the weights of Albert and Bella acting downwards. Then you also have the tension in the cable pulling the lift up.

Then start resolving. If you get stuck, please post all your working and diagrams.
9. (Original post by Muttley79)
Thanks - so what is the bit you are unclear about? I think you are confusing the forces acting on the whole system to those 'internal' forces acting on the people.
Yep that's exactly what happened. It feels like I have just had a moment of revelation. So what happened was I was mixing up the forces that was acting on the lift ONLY which is the lifts own mass, the reaction force coming from Albert and Bella and the tension, with the forces acting on the system as an whole where the forces on the whole system would be the the combined mass of the lift Albert and Bella and the tension. So what I can understand from this is that if I was working on the lift Only I would find the resultant force by doing T - total reaction force of the masses acting on the lift (From Newton's second law which states every action has a equal and opposite reaction, meaning if the lift exerts a force on Albert and Bella, Albert and Bella will exert a equal force on the lift in the opposite direction). And And equating the resultant force acting on the lift only to the mass of lift x acceleration to get a value for acceleration.

So if I was considering the system as a whole I would find the resultant force by doing T - Combined mass of lift Albert and Bella and equate it to the combined mass x acceleration to find the acceleration. But this would not be possible because we don't have the value for the masses.
Please correct me if I have gotten anything wrong. Thanks.
10. (Original post by notnek)
It really helps if you draw separate diagrams for lift questions. So draw 3 force diagrams for Albert, Bella and the system of the lift containing Albert and Bella.

Firstly, it's important to realise that all three of these objects will be moving with the same acceleration.

For Albert, the forces acting on him will be his weight and the force exerted on him by the lift. The forces acting on Bella will be similar.

For the lift containing Albert and Bella, you have the weights of the lift as well as the weights of Albert and Bella acting downwards. Then you also have the tension in the cable pulling the lift up.

Then start resolving. If you get stuck, please post all your working and diagrams.
We don't have wight of lift. Only the reaction force so we consider the lift individually and the forces acting upon it which is its own weight, the reaction force and the tension. Using the resultant force from this and equating it to mass of the lift times acceleration will give us the value for acceleration. Correct me if I'm wrong.
We don't have wight of lift. Only the reaction force so we consider the lift individually and the forces acting upon it which is its own weight, the reaction force and the tension. Using the resultant force from this and equating it to mass of the lift times acceleration will give us the value for acceleration. Correct me if I'm wrong.
You are given the mass of the lift as 250 so its weight is 250g.

When considering the lift, you should treat the lift, Albert and Bella as one system. So think of the system as a combined particle of mass (250 + mass of Albert + mass of Bella). Then you don't have to worry about the reaction forces.

EDIT : I should have tried the question.

So if I was considering the system as a whole I would find the resultant force by doing T - Combined mass of lift Albert and Bella and equate it to the combined mass x acceleration to find the acceleration. But this would not be possible because we don't have the value for the masses.
Please correct me if I have gotten anything wrong. Thanks.
I think you need to draw diagrams as Notnek suggests - we are told the forces the lift exerts so they are the ones we use for the 'whole system'.
13. (Original post by notnek)
You are given the mass of the lift as 250 so its weight is 250g.

When considering the lift, you should treat the lift, Albert and Bella as one system. So think of the system as a combined particle of mass (250 + mass of Albert + mass of Bella). Then you don't have to worry about the reaction forces.
I don't think you understand the question. If you look back at the question you are only provided with the reaction force of Albert and Bella and the mass of the lift. So you could only consider the lift separately.
14. (Original post by Muttley79)
I think you need to draw diagrams as Notnek suggests - we are told the forces the lift exerts so they are the ones we use for the 'whole system'.
Considering ring the whole system mean treating it as one particle therefore it requires the individual masses of Albert and Bella. Therefore you would have to consider the lift individually and the forces acting on it which includes the reaction forces Albert and Bella exert on the lift acting downwards and the tension acting upwards.
I don't think you understand the question. If you look back at the question you are only provided with the reaction force of Albert and Bella and the mass of the lift. So you could only consider the lift separately.
Yes the masses of Albert and Bella are unknowns but this is fine - just use variables. If you consider the lift completely separately then you're going to get confused.

In M1, for particles in the same straight line you can treat them as one system which is what you should be doing here. This is a technique that you need to get used to.

I strongly recommend that you draw three diagrams (one for Albert, one for Bella and one for the whole system) and post them here if you're unsure.

EDIT : I didn't try the question.
16. (Original post by notnek)
Yes the masses of Albert and Bella are unknowns but this is fine - just use variables. If you consider the lift completely separately then you're going to get confused.

In M1, for particles in the same straight line you can treat them as one system which is what you should be doing here. This is a technique that you need to get used to.

I strongly recommend that you draw three diagrams (one for Albert, one for Bella and one for the whole system) and post them here if you're unsure.
Part a of the question asks for the acceleration so setting up variables to work out the invidual masses (which is question part b and c) simultaneously would be highly inefficient. If found that considering the lift invidually first leads to significantly less working and gives answers immediately rather than having to to puzzle through many equations with unknown variables.
Part a of the question asks for the acceleration so setting up variables to work out the invidual masses (which is question part b and c) simultaneously would be highly inefficient. If found that considering the lift invidually first leads to significantly less working and gives answers immediately rather than having to to puzzle through many equations with unknown variables.
Okay I have to admit that I didn't try the question. I thought I recognised the question and considering the lift separately in the question I was thinking of would end up in a mess.

For this question it is more efficient to consider the lift separately, as you say. Sorry for the confusion - I'm trying to do my own work at the same time as helping but that's not a great excuse!

Can you please start again and explain why what you've been taught differs to the solution you're looking at?
18. (Original post by notnek)
If you got the correct answer by following your teachers method then that's fine. I'm not sure if you clarified whether you needed help because you got the wrong answer or wanted to understand the Solutionbank's answer.

If you're getting wrong answers in some questions using your teachers method then you should really think about doing them using the standard method I suggested. The algebra may be slightly harder (it's still quite simple once you've drawn the diagrams) but you're less likely to make a mistake from my experience.
I still use the standard method if I find that I could get to answer faster by using it, but my teachers method is what I have been using for a long time now and I feel like I could solve a lot of problems really fast using my teacher method because I have had so much practice with it. Anyways if my teacher does find me using your method he does throw mini tantrum which is why I am always cautious about using it.
I still use the standard method if I find that I could get to answer faster by using it, but my teachers method is what I have been using for a long time now and I feel like I could solve a lot of problems really fast using my teacher method because I have had so much practice with it. Anyways if my teacher does find me using your method he does throw mini tantrum which is why I am always cautious about using it.
You are completely correct in your approach to these questions. Sorry again.
20. (Original post by notnek)
You are completely correct in your approach to these questions. Sorry again.
Haha no worries.

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