Why Does the Normal Force Decrease When Pulling Upward on an Object?

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When pulling upward on an object, the normal force decreases because it counteracts both the gravitational force and the upward applied force. The correct expression for the normal force is N = mg - Fsin(θ), indicating that the upward force reduces the normal force rather than adding to it. Diagrams illustrating these forces are criticized for their clarity, as they can misrepresent the direction of the normal force. The normal force is a reaction force exerted by the surface, opposing the weight of the object. As the upward force increases, the normal force approaches zero when the object begins to lift off the surface.
David Donald
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I am trying to solve a homework problem that involves Work and friction but don't understand why if a force pulling upward on the object, the normal force becomes N = mg - Fsin(o)

shouldn't it be N = mg + Fsin(o) since both forces are acting upwards? (the normal force and the force applied)
 
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These diagrams are terrible. The normal force vector should point the opposite direction. The normal force is the force on the object exerted by the contact surface. The object is pulled downward by gravity, and in turn, the ground pushes upwards on the object. Notice that if you write it like that, you can add the vectors algebraically to get the net force, and then if you assume no acceleration, you should be able to see why the correct expression is N = mg - F \sin(\theta).
 
axmls said:
These diagrams are terrible. The normal force vector should point the opposite direction. The normal force is the force on the object exerted by the contact surface. The object is pulled downward by gravity, and in turn, the ground pushes upwards on the object. Notice that if you write it like that, you can add the vectors algebraically to get the net force, and then if you assume no acceleration, you should be able to see why the correct expression is N = mg - F \sin(\theta).
I think you are overstating the case. In general, a normal force is a compression at a boundary between objects. That is, there is a pair of equal and opposite forces. Which you consider to be 'the' normal depends on your perspective. As far as the ground is concerned, the normal force is as shown in the diagrams.
That said, drawing F as a force applied to the block but N as a force the block applies to the ground is certainly inconsistent. A free body diagram should show all the forces from the same point of view.
 
David Donald said:
I am trying to solve a homework problem that involves Work and friction but don't understand why if a force pulling upward on the object, the normal force becomes N = mg - Fsin(o)

I agree the diagrams are poor.

Imagine if the applied force F was vertical (pulling upwards) and you increased F slowly from zero until the object just lifted off the surface. The Normal force would reduce and reach zero as the block lifts off.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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