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admierewebb
14.09.2019 •
Engineering
Acarpenter 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?
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Ответ:
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
Ответ: