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    Just a rather basic question which I seem to be stuck on..

    Q. A man of mass 75kg travels upward in a lift which is decelerating at 0.4ms^-2. Find the force exerted by the man on the floor of the lift.

    So thinking about a force diagram, there are two forces;
    1. The weight of the man acting downward on the floor of the lift, having magnitude 75x9.8=735N.
    2. The normal reaction force, R, from the lift floor on the man. In order for the lift to be decelerating with magnitude 0.4ms^-2 upward then W-R=ma which gives 735-R=75(0.4) giving R=705N.

    So the upward force (normal contact force) on the man is 705N and the downward force (the weight) is 735N. So, as the question asks for, the force exerted by the man on the floor then surely this is the weight of the man exerted on the floor ie 735N. The solution however is 705N but this is the force acting on the man from the floor not the reverse...why?
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    (Original post by Mathematicus65)
    Just a rather basic question which I seem to be stuck on..

    Q. A man of mass 75kg travels upward in a lift which is decelerating at 0.4ms^-2. Find the force exerted by the man on the floor of the lift.

    So thinking about a force diagram, there are two forces;
    1. The weight of the man acting downward on the floor of the lift, having magnitude 75x9.8=735N.
    2. The normal reaction force, R, from the lift floor on the man. In order for the lift to be decelerating with magnitude 0.4ms^-2 upward then W-R=ma which gives 735-R=75(0.4) giving R=705N.

    So the upward force (normal contact force) on the man is 705N and the downward force (the weight) is 735N. So, as the question asks for, the force exerted by the man on the floor then surely this is the weight of the man exerted on the floor ie 735N. The solution however is 705N but this is the force acting on the man from the floor not the reverse...why?
    I got 705 N. Note: decelerating. It seems like you've considered the magnitude of acceleration to be positive in your calculation!
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    (Original post by Mathematicus65)
    Just a rather basic question which I seem to be stuck on..

    Q. A man of mass 75kg travels upward in a lift which is decelerating at 0.4ms^-2. Find the force exerted by the man on the floor of the lift.

    So thinking about a force diagram, there are two forces;
    1. The weight of the man acting downward on the floor of the lift, having magnitude 75x9.8=735N.
    2. The normal reaction force, R, from the lift floor on the man. In order for the lift to be decelerating with magnitude 0.4ms^-2 upward then W-R=ma which gives 735-R=75(0.4) giving R=705N.

    So the upward force (normal contact force) on the man is 705N and the downward force (the weight) is 735N. So, as the question asks for, the force exerted by the man on the floor then surely this is the weight of the man exerted on the floor ie 735N. The solution however is 705N but this is the force acting on the man from the floor not the reverse...why?
    What you are really doing is finding the reaction force. Because the reaction force is the force (contact force) that the floor of the lift exerts on the man, is equal to the force that the man exerts on the floor of the lift. So really the question is find the reaction force but it would not be testing your understanding as well then.
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    (Original post by Mathematicus65)
    Just a rather basic question which I seem to be stuck on..

    Q. A man of mass 75kg travels upward in a lift which is decelerating at 0.4ms^-2. Find the force exerted by the man on the floor of the lift.

    So thinking about a force diagram, there are two forces;
    1. The weight of the man acting downward on the floor of the lift, having magnitude 75x9.8=735N.
    2. The normal reaction force, R, from the lift floor on the man. In order for the lift to be decelerating with magnitude 0.4ms^-2 upward then W-R=ma which gives 735-R=75(0.4) giving R=705N.

    So the upward force (normal contact force) on the man is 705N and the downward force (the weight) is 735N. So, as the question asks for, the force exerted by the man on the floor then surely this is the weight of the man exerted on the floor ie 735N. The solution however is 705N but this is the force acting on the man from the floor not the reverse...why?
    remember when you use F=ma the resultant force F goes in the direction of the acceleration so

    Acceleration is -0.4ms^-2 upwards
    mass of man is 75g downwards
    normal reaction going up from the man is R
    F=ma
    R-75g=75x-0.4
    find R and you have the force exerted by the man on the floor of the lift
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    (Original post by aymanzayedmannan)
    I got 705 N. Note: decelerating. It seems like you've considered the magnitude of acceleration to be positive in your calculation!
    It should generate the same solution. The lift is travelling upward but is decelerating. This tells us that the downward force (the weight of the man) is greater than the reaction force from the floor on the man right? Therefore if taking upward to be positive then

    R - 75g = 75 x -0.4
    R = 705 N (the same answer I originally obtained by considering motion downwards since a deceleration upward is the same as an acceleration downward)

    My concern is why the force exerted by the man on the floor is the normal contact force? The normal contact force is the force exerted by the floor on the man. So surely the force exerted by the man on the floor is the man's weight, not the normal contact force. Hence being 735 N?
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    (Original post by B_9710)
    What you are really doing is finding the reaction force. Because the reaction force is the force (contact force) that the floor of the lift exerts on the man, is equal to the force that the man exerts on the floor of the lift. So really the question is find the reaction force but it would not be testing your understanding as well then.

    If the reaction force exerted on the man by the lift floor is equal to the force that the man exerts on the floor of the lift, then surely the lift would have constant upward velocity or be stationary since the forces acting upon it are in equilibrium. In order to produce a deceleration the downward force surely must be greater than the upward force ie weight of the man is greater than the normal reaction force from the floor? Hence the force acting on the floor by the man must be his weight?
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    (Original post by Mathematicus65)
    It should generate the same solution. The lift is travelling upward but is decelerating. This tells us that the downward force (the weight of the man) is greater than the reaction force from the floor on the man right? Therefore if taking upward to be positive then

    R - 75g = 75 x -0.4
    R = 705 N (the same answer I originally obtained by considering motion downwards since a deceleration upward is the same as an acceleration downward)

    My concern is why the force exerted by the man on the floor is the normal contact force? The normal contact force is the force exerted by the floor on the man. So surely the force exerted by the man on the floor is the man's weight, not the normal contact force. Hence being 735 N?
    Ah yes, my mistake. I thought you were getting 735 from your calculation. The force exerted on the floor is indeed the weight of the man, whereas the force exerted by the floor on the man is the normal reaction which is calculated using N2L.
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    (Original post by Mathematicus65)
    My concern is why the force exerted by the man on the floor is the normal contact force? The normal contact force is the force exerted by the floor on the man. So surely the force exerted by the man on the floor is the man's weight, not the normal contact force. Hence being 735 N?
    Are you familiar with Newton's third law? The force vector exerted on the man by the floor is of equal magnitude to the force vector exerted by the floor on the man, but it points in the opposite direction i.e. the two vectors are of equal length, but are anti-parallel.
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    (Original post by thefatone)
    remember when you use F=ma the resultant force F goes in the direction of the acceleration so

    Acceleration is -0.4ms^-2 upwards
    mass of man is 75g downwards
    normal reaction going up from the man is R
    F=ma
    R-75g=75x-0.4
    find R and you have the force exerted by the man on the floor of the lift
    A deceleration of 0.4ms^-2 upward is the same as an acceleration of 0.4ms^-2 downward. I just considered the motion downward, and hence why I used 0.4ms^-2 in the calculation instead of -0.4ms^-2. It generates the same answer:

    75g-R=75 x 0.4
    R = 705 N

    I am concerned with the original question - which asked what is the force exerted by the man on the floor. The normal contact force acts upward. It is the force exerted by the floor on the man (the reverse of what the question asked for). Hence surely the force exerted by the man on the floor is the man's weight. W=75 x 9.8 = 735N...?
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    (Original post by aymanzayedmannan)
    The force exerted on the floor is indeed the weight of the man
    The weight of the man is the gravitational force exerted on the man by the Earth. It's not directly related to the force exerted on whichever surface that he is standing on.
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    (Original post by atsruser)
    Are you familiar with Newton's third law? The force vector exerted on the man by the floor is of equal magnitude to the force vector exerted by the floor on the man, but it points in the opposite direction i.e. the two vectors are of equal length, but are anti-parallel.

    Yes I understand Newton's Third Law. But assuming that the force exerted on the floor by the man was equal in magnitude to the force exerted by the man on the floor, then the man would be in equilibrium; the forces on him would be balanced. And hence by Newton's First Law of Mechanical Motion then the man would have constant velocity or be stationary since a net force is required for an acceleration or deceleration...? Since the lift is moving with a deceleration of 0.4ms^-2 the forces on him must be unbalanced. ie The weight is greater than the reaction force... So why is the force exerted by the man on the floor given by the normal contact force, rather than his weight?
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    (Original post by Mathematicus65)
    Yes I understand Newton's Third Law. But assuming that the force exerted on the floor by the man was equal in magnitude to the force exerted by the man on the floor, then the man would be in equilibrium; the forces on him would be balanced.
    No. The force exerted on the man by the floor, and the force exerted on the floor by the man act on *different* bodies so they can't be added together, anymore than my weight right now can be meaningfully added to your weight right now - to find the motion of a body, you need to consider the forces acting on that body, and the force exerted by the man on the floor is not a force on the man. This is why we draw free body diagrams, so we can clearly isolate the forces, then apply Newton II correctly.
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    (Original post by atsruser)
    The weight of the man is the gravitational force exerted on the man by the Earth. It's not directly related to the force exerted on whichever surface that he is standing on.

    Oops, my bad. I was correct initially it seems.

    (Original post by Mathematicus65)
    So why is the force exerted by the man on the floor given by the normal contact force, rather than his weight?
    Please read Atsruser's reply to my post.
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    The gravitational force on the man from the Earth is the same as the gravitational force that the man exerts on the Earth. So the man is causing the Earth to accelerate towards him by Newtons 3rd law.
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    (Original post by Mathematicus65)
    Just a rather basic question which I seem to be stuck on..

    Q. A man of mass 75kg travels upward in a lift which is decelerating at 0.4ms^-2. Find the force exerted by the man on the floor of the lift.

    So thinking about a force diagram, there are two forces;
    1. The weight of the man acting downward on the floor of the lift, having magnitude 75x9.8=735N.
    2. The normal reaction force, R, from the lift floor on the man. In order for the lift to be decelerating with magnitude 0.4ms^-2 upward then W-R=ma which gives 735-R=75(0.4) giving R=705N.

    So the upward force (normal contact force) on the man is 705N and the downward force (the weight) is 735N. So, as the question asks for, the force exerted by the man on the floor then surely this is the weight of the man exerted on the floor ie 735N. The solution however is 705N but this is the force acting on the man from the floor not the reverse...why?
    The weight of the man acts on the man, not on anything else.

    Go out and find a lift that isn't too gentle. You will find the force on your feet from the floor varies. If you are going upwards, the force on your feet will be bigger at the beginning of the journey when you and the lift are accelerating. When the lift decelerates at the end of the journey, the force will be smaller. Meanwhile the force of your feet pushing on the floor will be equal and in the opposite direction to the force of the floor pushing on your feet.
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    (Original post by Mathematicus65)
    A deceleration of 0.4ms^-2 upward is the same as an acceleration of 0.4ms^-2 downward. I just considered the motion downward, and hence why I used 0.4ms^-2 in the calculation instead of -0.4ms^-2. It generates the same answer:

    75g-R=75 x 0.4
    R = 705 N

    I am concerned with the original question - which asked what is the force exerted by the man on the floor. The normal contact force acts upward. It is the force exerted by the floor on the man (the reverse of what the question asked for). Hence surely the force exerted by the man on the floor is the man's weight. W=75 x 9.8 = 735N...?
    Force exerted by the man on the floor isn't his weight. Weight is the force man exerts upon the Earth.

    Force exerted on the floor by the person=Force exerted on person by floor
    Due to Newtons 3rd Law
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    This thread is an example of a common problem for students in M1.

    You often have R = mg in M1 problems when a particle is moving along a horizontal ground. For many students, this and N3 law leads to the thinking that the normal reaction will always be equal to the weight.

    This student then gets a motion in a lift question which blows their mind
 
 
 
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