A carpenter uses a hammer to strike a nail. Approximate the hammer's weight of 1.8lbs, as being concentrated at the head, and assume that at impact the head is traveling in the - j direction. If the hammer contacts the nail at 50mph and the impact occurs over 0.023 seconds, what is the magnitude of the average force exerted by the nail of the hammer?

The average force F exherted by the nail over the hammer is 178.4 lbf.

Explanation:

The force F exherted by the nail over the hammer is defined as:

F = |I|/Δt

Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:

I = ΔP = Pf - Pi

Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:

P = m. v

Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:

ΔP = - Pi = - m. vi

The initial velocity is given as 50 mph and we will expressed in ft/s:

vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s

By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:

|I| = |ΔP| = |-m. vi| = 1.8 lb * 73.3 ft/s = 132 lb. ft/s

Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:

F = 132 lb. ft/s / 0.023 s = 5740 lb. ft/s^2

Expressing in lbf:

F = 5740 lb. ft/s^2 * 0.031 lbf/lb. ft/s^2 = 178.4 lbf