This chapter's a sadist. please see this question. A drop-forge hammer, of mass 1500kg, falls under gravity...
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This chapter's a sadist. please see this question. A drop-forge hammer, of mass 1500kg, falls under gravity...

[From: ] [author: ] [Date: 12-09-04] [Hit: ]
Hence force exerted = 15000-2250=12750N ??????......
onto a piece of hot metal which rests in a fixed die. From the instant that the hammer strikes the piece of metal until it comes to rest, the hammer is decelerating at 1.5 m/s^2. Make sketches showing the forces on the hammer and on the metal while they are in contact.

Find the magnitude of the force exerted by the hammer on the piece of metal

a) while the hammer is decelerating

b) after the hammer has come to rest?

What do I do? Why isn’t it like:
Force exerted by hammer on metal = weight – force acting on it in opposite direction (decelerating it)
The weight = 15000N.
The force that acts in opposite direction and decelerates it is 1500(1.5)=2250N.
Hence force exerted = 15000-2250=12750N ????????
This is the concept I have been following all this while in all questions before this doomed chapter of newton’s third law. Isn’t it obvious that a force that acts upward decelerating it needs to be subtracted from weight to get net force acting downward on hammer?

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First do a force diagram. Then find the values. Don't try to find the value of a force and then put it on your force diagram.

Or put another way, forces arise from causes. You are trying to create a force from an effect, the deceleration. There is no such force as "the deceleration force".

There are two forces on the hammer: weight and support force. They act in opposite directions. I like to make up positive so the weight will subtract.

Fhammer = S - W

Now the acceleration of the hammer is the result of the TOTAL force on the hammer. So

Fhammer = S - W = ma

You were trying to start out with S = ma

Be careful to get the right sign on a. The hammer is moving downward, which is negative, but slowing down, so the sign of the acceleration is opposite the velocity. Therefore a is positive. And therefore S - W > 0 and S > W.

You might be more comfortable having positive be the direction things are moving. Then a deceleration is always a negative acceleration. That's ok, but don't get too attached to it. It will mess you up when you get a problem where something changes direction partway through the problem.
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