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    Albert and Bella are both standing In a lift, mass of lift is 250N, as lift moves upward with constant acceleration the floor of the lift exerts force 678N on Albert and force 452 on Bella,the tension in the cable pulling the lift upwards is 3955N, find the acceleration of the lift.

    I get the method of how to solve but in one diagram the forces the lift exerts and on Bella and Albert are pointed down instead of up from the floor, I don't get this I thought when a object is accelerating the weight and normal reaction force are not equal .
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    (Original post by uchihaitachi0023)
    Albert and Bella are both standing In a lift, mass of lift is 250N, as lift moves upward with constant acceleration the floor of the lift exerts force 678N on Albert and force 452 on Bella,the tension in the cable pulling the lift upwards is 3955N, find the acceleration of the lift.

    I get the method of how to solve but in one diagram the forces the lift exerts and on Bella and Albert are pointed down instead of up from the floor, I don't get this I thought when a object is accelerating the weight and normal reaction force are not equal .
    Can you please post a picture showing this?
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    Name:  IMG_0129.PNG
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    (Original post by notnek)
    Can you please post a picture showing this?
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    (Original post by uchihaitachi0023)
    ...
    This is a diagram of all the forces acting on the lift only. Albert and Bella exert a force on the lift pointing downwards, which is why the arrows are pointing down.

    Force diagrams will always show the forces acting on a single object (or possibly some combined objects). In this case the diagram is only showing forces acting on the lift so e.g. the forces that act upwards on Albert and Bella won't be shown on this diagram.

    If you were to draw a diagram of the forces acting on Albert then you would have a down arrow showing Albert's weight and an up arrow showing the force exerted on Albert by the lift.
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    But why are the forces exerted from the floor into them equal to the force they exert onto the floor, I thought Newton's third law only applies to objects that are not accelerating, but aren't they accelerating as the lift accelerates?



    (Original post by notnek)
    This is a diagram of all the forces acting on the lift only. Albert and Bella exert a force on the lift pointing downwards, which is why the arrows are pointing down.

    Force diagrams will always show the forces acting on a single object (or possibly some combined objects). In this case the diagram is only showing forces acting on the lift so e.g. the forces that act upwards on Albert and Bella won't be shown on this diagram.

    If you were to draw a diagram of the forces acting on Albert then you would have a down arrow showing Albert's weight and an up arrow showing the force exerted on Albert by the lift.
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    (Original post by uchihaitachi0023)
    But why are the forces exerted from the floor into them equal to the force they exert onto the floor, I thought Newton's third law only applies to objects that are not accelerating, but aren't they accelerating as the lift accelerates?
    No Newtons's Third Law always applies.

    The law says that if A exerts a force on B then B will exert an equal and opposite force on A.

    Albert exerts a force on the lift (not necessarily equal to Albert's weight - this is very important) and so the lift exerts an equal force in the opposite direction on Albert.

    If the lift wasn't accelerating then the force that Albert exerted on the lift would be equal in magnitude to his weight. This isn't the case if the lift is accelerating.

    The important point is that Albert's weight and the force that he exerts on the lift are two different things. Albert's weight is the force exerted on Albert by the Earth and this is not necessarly equal in magnitude to the force exerted by Albert on the lift.

    But it will always be the case by Newton's Third Law that the force exerted by Albert on the lift will be equal in magnitude to the force that the lift exerts on Albert.

    This is a concept that M1 students often find very hard to understand so don't worry if you're still unsure. Please continue to ask questions if anything doesn't make sense.
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    (Original post by notnek)
    No Newtons's Third Law always applies.

    The law says that if A exerts a force on B then B will exert an equal and opposite force on A.

    Albert exerts a force on the lift (not necessarily equal to Albert's weight - this is very important) and so the lift exerts an equal force in the opposite direction on Albert.

    If the lift wasn't accelerating then the force that Albert exerted on the lift would be equal in magnitude to his weight. This isn't the case if the lift is accelerating.

    The important point is that Albert's weight and the force that he exerts on the lift are two different things. Albert's weight is the force exerted on Albert by the Earth and this is not necessarly equal in magnitude to the force exerted by Albert on the lift.

    But it will always be the case by Newton's Third Law that the force exerted by Albert on the lift will be equal in magnitude to the force that the lift exerts on Albert.

    This is a concept that M1 students often find very hard to understand so don't worry if you're still unsure. Please continue to ask questions if anything doesn't make sense.
    Ohh okay my problem was thinking the force he exerts onto the floor is equal to his weight, I get it now, one problem tho why aren't they the same I mean what other type of force can he exert other than weight if he's just standing?
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    (Original post by uchihaitachi0023)
    Ohh okay my problem was thinking the force he exerts onto the floor is equal to his weight, I get it now, one problem tho why aren't they the same I mean what other type of force can he exert other than weight if he's just standing?
    You're not the only one who thought weight and contact force are the same thing - I've got a feeling that over half of M1 students also think this.

    The weight of Albert is the force pulling Albert towards the centre of the Earth. This is different to the contact force that Albert's feet apply to the surface. They may have the same magnitude but they are not the same thing.

    You're so used to horizontal stationary surfaces in M1 where the contact force on the surface is the same magnitude as the weight, which is why you're confused. But try to imagine you were in a lift and the cable broke and you accelerated fast downwards.The force that your feet applied to the floor of the lift would be much less than if the lift was stationary and eventually you'd feel "weightless" where your feet exerted practically no force on the lift. Can you imagine this?

    (By the way don't let the term 'weightless' confuse you - the force pulling you towards the Earth remains the same).

    Going back to the M1 topic, the most important thing to do in lift questions is to realise before you do anything that the force that the person exerts on the lift is an unknown. And this also means that the force exerted by the lift on the person is unknown, since they have the same magnitude by Newton's Third Law. You have to do calculations to work out what this force is.
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    Thank you you've been great help especially at this time, I would of probably attempted this for hours without even grasping the correct concept to solve it

    (Original post by notnek)
    You're not the only one who thought weight and contact force are the same thing - I've got a feeling that over half of M1 students also think this.

    The weight of Albert is the force pulling Albert towards the centre of the Earth. This is different to the contact force that Albert's feet apply to the surface. They may have the same magnitude but they are not the same thing.

    You're so used to horizontal stationary surfaces in M1 where the contact force on the surface is the same magnitude as the weight, which is why you're confused. But try to imagine you were in a lift and the cable broke and you accelerated fast downwards.The force that your feet applied to the floor of the lift would be much less than if the lift was stationary and eventually you'd feel "weightless" where your feet exerted practically no force on the lift. Can you imagine this?

    (By the way don't let the term 'weightless' confuse you - the force pulling you towards the Earth remains the same).

    Going back to the M1 topic, the most important thing to do in lift questions is to realise before you do anything that the force that the person exerts on the lift is an unknown. And this also means that the force exerted by the lift on the person is unknown, since they have the same magnitude by Newton's Third Law. You have to do calculations to work out what this force is.
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    (Original post by uchihaitachi0023)
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    Notnek
    I'm a little confused with this one.

    The forces acting on the lift are 3955 - (678 + 452 + 250g)
    So F = ma
    3955 - (678 +452 + 250g) = ma

    But here it says that the mass is equal to 250?
    When isn't the mass equal to 250 + A + B?
    Or is it because we've already taken into the account the force that A & B exert on the lift we don't need to worry about their mass?
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    (Original post by Retsek)
    Notnek
    I'm a little confused with this one.

    The forces acting on the lift are 3955 - (678 + 452 + 250g)
    So F = ma
    3955 - (678 +452 + 250g) = ma

    But here it says that the mass is equal to 250?
    When isn't the mass equal to 250 + A + B?
    Or is it because we've already taken into the account the force that A & B exert on the lift we don't need to worry about their mass?
    3955 - (678 +452 + 250g) is the resultant force acting on the lift only.

    If you were considering the whole system of the lift, A and B then the downwards force would need to be the the total weight of the lift, A and B but the weights of A and B are unknown so you can't. If you were considering the whole system you would have a resultant force of

    3955 - (250 + A + B)g

    There would be no need to include the 678 and 452 forces because these are just forces acting on parts of the whole system.

    So back to my original point, since you have the resultant force acting on the lift only, when you use F = ma, the m has to be the mass of the lift only.
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    (Original post by Notnek)
    3955 - (678 +452 + 250g) is the resultant force acting on the lift only.

    If you were considering the whole system of the lift, A and B then the downwards force would need to be the the total weight of the lift, A and B but the weights of A and B are unknown so you can't. If you were considering the whole system you would have a resultant force of

    3955 - (250 + A + B)g

    There would be no need to include the 678 and 452 forces because these are just forces acting on parts of the whole system.

    So back to my original point, since you have the resultant force acting on the lift only, when you use F = ma, the m has to be the mass of the lift only.
    The question has been the bane of a lot of people's textbook grind.

    I hope you don't mind answering some more questions regarding this.

    Q1: In Part (a) of this question, the acceleration of the lift was found, without the two people inside (without their weights being considered). But in the next part of the question, it was used that the acceleration of Albert was  1.5ms^{-2} . Surely, this cannot be correct, as we didn't we even consider the weights of the two people inside the lift for part (a). The way I'm thinking about it is in part (a) the acceleration of the lift was found WITHOUT two people inside which in my opinion would not equal the acceleration of the lift WITH the two people inside.

    Please consider this. Suppose we have a ball of mass 0.2kg that is attached to the upper end of a verticle light rod. The thrust in the rod when it raises the ball is  2.3N . Find the acceleration. So, the way I am going to solve this problem is comparable to how the above problem was solved. We have our thrust force pushing the ball up with a force of  2.3 N and we have our force due to gravity of  0.2 \times 9.8 N .

     2.3 N = 0.2a

     a = \frac{2.3N}{0.2kg}

    However, clearly I have forgotten the force of thrust and so this incorrect. But why is this not analogous to what was done in the above question???
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    (Original post by MathQS)
    The question has been the bane of a lot of people's textbook grind.

    I hope you don't mind answering some more questions regarding this.

    Q1: In Part (a) of this question, the acceleration of the lift was found, without the two people inside (without their weights being considered). But in the next part of the question, it was used that the acceleration of Albert was  1.5ms^{-2} . Surely, this cannot be correct, as we didn't we even consider the weights of the two people inside the lift for part (a). The way I'm thinking about it is in part (a) the acceleration of the lift was found WITHOUT two people inside which in my opinion would not equal the acceleration of the lift WITH the two people inside.

    Please consider this. Suppose we have a ball of mass 0.2kg that is attached to the upper end of a verticle light rod. The thrust in the rod when it raises the ball is  2.3N . Find the acceleration. So, the way I am going to solve this problem is comparable to how the above problem was solved. We have our thrust force pushing the ball up with a force of  2.3 N and we have our force due to gravity of  0.2 \times 9.8 N .

     2.3 N = 0.2a

     a = \frac{2.3N}{0.2kg}

    However, clearly I have forgotten the force of thrust and so this incorrect. But why is this not analogous to what was done in the above question???
    I just realised I didn't reply to this but it's going to take a bit of reading to understand what your question is! Can you please remind me to reply tomorrow or when you're next free if I haven't replied by then?
 
 
 
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