How Does a Pile Driver Generate a Normal Force Greater Than Its Weight?

[Cyrus]

I was looking at a problem involving a pile driver. It shows that the normal force is greater than the force due to the weight. I was wondering how the normal force manages to be large than the force due to the weight.

a.) If the pile driver is in free fall, will its force at any instance during the free fall be F=ma?

b.) if a.) is true, then should it not deliver a force of F=ma on the beam it hits, and in turn, the beam have a normal force equal and opposite to the weight?


c.) if the normal force is larger than the force due to the weight, then is the action not equal and opposite to the reaction? It hits with a force mg, but reacts with a normal force much larger.

Any help krab? your an expert at everything

 

[russ_watters]

When in freefall, it accelerates for a second or two - how long does it decelerate for? If the acceleration takes 2 seconds (for example) and the deceleration takes .01 sec, how much more force is involved...?

 

[Cyrus]

Could you explain that again please. I see what you mean. If it decelerates much faster than it acelerated, then the force will be larger. (this would corespond to a larger normal force?). But would that not mean that the weight vector should be equaly big but in the oposite direction to the normal? How is the normal still bigger than the weight vector.


W_t_o_t = (w-f-n)s

here they use the following equation to find the normal force during the impact. S is the distance it takes to come to a stop. If it decelerates for a faster amount of time, then wouldent the weight be equal and opposite to the normal for that amount of time? In the equation the weight w is just the weight F=m*9.81.

 

[Cyrus]

Hmm, looking at it a second time, there must be some force opposite to the direction of motion, or the thing would have no way of stopping! If the normal force were equal and opposite to the weight, there would be a balance of forces, and it would just fall forever! and that can't be right. So the larger N must be the force that provides the deceleration of the pile driver. This is very counter intuitive to me, It seems that we have an action, with a greater reaction.

 

[Cyrus]

Ah, perhaps there is an equal and opposite force after all. The normal force is not the action reaction pair with the weight vector. The normal force is what is used to bring it to a hault. There is an action reaction pair, and that force IS equal and opposite to the normal, but it is experience by the I beam. In the physics book their free body diagram has an upward normal, an upward frictional force, and a downward weight. I was thinking of the downward weight as the action force, and the upward normal as the reaction force. But this is incorrect. The reaction was not displayed on the free body diagram. It too should be equal in magintude to the normal.

 

[russ_watters]

Just to clarify, when you say "normal force", you mean the net force between the pile and the driver? Weight is weight either way (its constant, of course) and the rest of the force is the force decelerating the driver.

There are 3 forces here:

-weight of driver
-deceleration force
-net force (weight + deceleration force)