Why is it harder to hammer in a nail the deeper it gets?

  • Thread starter Thread starter mrneuro
  • Start date Start date
AI Thread Summary
As a nail is hammered deeper into wood, the friction between the nail and the wood increases due to the expanding contact area, which leads to a greater normal force acting on the nail. This normal force is not constant; it increases as more of the nail is embedded, resulting in greater resistance to further penetration. The discussion highlights that the relationship between pressure and area differs for a nail compared to other objects, as the normal force is influenced by the wood's response to being compressed. Consequently, the difficulty in hammering the nail increases with depth due to this rising friction and normal force. Understanding these dynamics clarifies why deeper penetration requires more effort.
mrneuro
Messages
5
Reaction score
0
Why is it harder to hammer in a nail the deeper it gets?

Classical physics says

F k or s = μ k or s N

the normal force is constant right? or is the nail's normal force increasing as it enters the piece of wood?

or is it one of those exceptions like why wider tires on snow work better than thinner ones? (b/c of pressure exertion over an area)

could someone please explain this its not making any sense
 
Physics news on Phys.org
Think qualitatively for a moment. As the nail gets deeper in the material, what has changed?
 
There's more friction with the wood on the sides of the nail.
 
bcbwilla said:
Think qualitatively for a moment. As the nail gets deeper in the material, what has changed?

the pressure exerted by the wood onto the nail in this case the normal force is increasing?

is that why? was i just overlooking the fact that N is increasing in this case and N ≠ mg
 
The normal force (pushing radially inwards) is indeed increasing as more and more of the nail is gripped by the wood.

Whereas, to a very rough approximation, the area of contact doesn't matter for the frictional force between a body and the surface it's resting on, because the pressure goes down if the area goes up, there's no such effect with the nail, because the normal contact force isn't equal to a fixed weight but arises in response to the wood being strained.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top