Hammer + Nail forces Will answer in response

[sadakaa]

1. A hammer is pounding a nail straight down into a wooden board. Just before the hammer hits the nail, the speed of the hammer is 1.2 m/s. The hammer drives the nail into the wood and stops after the nail goes in a distance of 0.95 cm. The weight of the hammer head is 6N, and in addition to this weight there is a force of 14 N exerted on the hammer head by the person using the hammer. Assume that the acceleration of the hammerhead is constant while it is contact with the nail and moving downward, (i.e., while the hammer is slowing from its initial speed to zero). Calculate the magnitude of the downward force exerted by the hammer head on the nail while it is driving the nail into the wood. (Hint: By Newton’s third law, the force of the hammer head on the nail has equal magnitude to the force of the nail up on the hammer head. Ask yourself what are all the forces on the hammer head.)

Homework Equations

V² = V₀² + 2a(x-x₀)

The Attempt at a Solution

Take up as positive, thus:

V0 = -1.2 m/s
V = 0
X0 = .0095 m
X = 0

0 = -1.2² +2(a)(-.0095)
a = 75.7895 m/s²

Now i get stuck. I set up the equation

ma = F_normal - F_person - F_grav

I'm not too sure how to continue. Help!

 

[LowlyPion]

1. A hammer is pounding a nail straight down into a wooden board. Just before the hammer hits the nail, the speed of the hammer is 1.2 m/s. The hammer drives the nail into the wood and stops after the nail goes in a distance of 0.95 cm. The weight of the hammer head is 6N, and in addition to this weight there is a force of 14 N exerted on the hammer head by the person using the hammer. Assume that the acceleration of the hammerhead is constant while it is contact with the nail and moving downward, (i.e., while the hammer is slowing from its initial speed to zero). Calculate the magnitude of the downward force exerted by the hammer head on the nail while it is driving the nail into the wood. (Hint: By Newton’s third law, the force of the hammer head on the nail has equal magnitude to the force of the nail up on the hammer head. Ask yourself what are all the forces on the hammer head.)

Homework Equations

V² = V₀² + 2a(x-x₀)

The Attempt at a Solution

Take up as positive, thus:

V0 = -1.2 m/s
V = 0
X0 = .0095 m
X = 0

0 = -1.2² +2(a)(-.0095)
a = 75.7895 m/s²

Now i get stuck. I set up the equation

ma = F_normal - F_person - F_grav

I'm not too sure how to continue. Help!

Looks like you are almost there.

You have the nail pushing up with the net deceleration on the mass of the hammer head and the force of the arm and shoulder being present throughout the deceleration as well as gravity on the mass. So I would write it as

F = (75.79)*(6N)/(9.8) +14N + 6N

 

[baba89]

Based on the given information, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the object is the hammer head and the net force is the sum of all the forces acting on it.

So, we can write the equation as:

ma = Fnet

Where m is the mass of the hammer head, a is the acceleration, and Fnet is the net force acting on the hammer head. Let's break down the forces acting on the hammer head:

1) Weight of the hammer head (6N)
2) Force exerted by the person (14N)
3) Normal force from the wood (Fnormal)
4) Force of the nail on the hammer head (Fnail)

Since the hammer head is moving downward and slowing down, the net force will be in the upward direction. So, we can write:

Fnet = Fnormal + Fnail - Fperson - Fgrav

Now, we know that the force of the nail on the hammer head (Fnail) is equal in magnitude to the force of the hammer head on the nail (by Newton's third law). So, we can rewrite the equation as:

ma = Fnormal + Fnormal - Fperson - Fgrav

Simplifying it further:

ma = 2Fnormal - Fperson - Fgrav

Now, we need to find the value of Fnormal. We can use the equation of motion to find the acceleration (a) and then substitute it in the above equation to solve for Fnormal.

From the given information, we know that the hammer head stops after traveling a distance of 0.95 cm. So, we can write:

V2 = V02 + 2a(x-x0)

Where V0 is the initial velocity (1.2 m/s), V is the final velocity (0 m/s), x0 is the initial position (0 m), and x is the final position (0.0095 m).

Substituting the values, we get:

0 = (1.2)2 + 2a(0.0095)

Solving for a, we get:

a = -75.7895 m/s2

Substituting this value in the previous equation, we get:

m(-75.7895) = 2Fnormal - 14 - 6

Solving for F