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: ]
and the acceleration, so we can solve for the decelerating force.Weight/gravity is down (positive), acceleration is up (negative)F(decelerating force) = (m*a) - F(weight) = (m * a) - (m * g)F(dec) = (1500kg * (-1.5m/s^2)) - (1500kg * 10m/s^2)F(dec) = 1500kg * (-1.5m/s^2 - 10m/s^2)F(dec) = 1500kg * (-11.......

You seem to be asking a lot of questions about the hammer, so lets start there.

(a) during deceleration. (I'll use down for positive and up for negative).
F(net) = F(weight) + F (declerating force) = m * a

We know the weight, the mass, and the acceleration, so we can solve for the decelerating force.
Weight/gravity is down (positive), acceleration is up (negative)
F(decelerating force) = (m*a) - F(weight) = (m * a) - (m * g)
F(dec) = (1500kg * (-1.5m/s^2)) - (1500kg * 10m/s^2)
F(dec) = 1500kg * (-1.5m/s^2 - 10m/s^2)
F(dec) = 1500kg * (-11.5m/s^2)
F(dec) = -17250N

The force the die is pushing on the hammer (during the deceleration) is 17250N upward. Therefore the hammer must be pushing 17250N downward on the die.

Now for (b) hammer is still.
F(net) = F(die) + F(weight) = m*a
F(die) = m*a - F(weight)
F(die) = 0 - 15000N
F(die) = -15000N
So while resting, the die is exactly resisting the weight of the hammer by providing a force of 15000N upward.

Does that help?
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