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Force exerted by a seatbelt, on driver, after collision

  • Thread starter Iciris.0
  • Start date
  • #1
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Homework Statement



Suppose that the car is stopped from 60mph within right after the collision and the driver (75kg) didn’t push the brake.

Seatbelt is non-stretchable and produces a constant force to stop the driver in 10cm. Calculate the acceleration of the driver and the force that the seatbelt acts on the driver for above two conditions.


Homework Equations



Time for the driver to hit the seatbelt: t=d/v
Force that the seatbelt exerts on the driver: F=m(vo-vt)/t
Acceleration of the driver: a=F/m


The Attempt at a Solution



60mph=27m/s

Time for the driver to hit the seatbelt:
t=d/v = 0.1m/(27m/s) = 3.7ms

Force that the seatbelt exerts on the driver:
F=m(vo-vt)/t = 75kg(27m/s - 0m/s)/3.7ms = 547,297N

Acceleration of the driver:
a=F/m = 547,297N/75kg = 7,297.3 m/s^2

Is this right, the force seems so high?
 

Answers and Replies

  • #2
CWatters
Science Advisor
Homework Helper
Gold Member
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2,295
You appear to be out by a factor of 2...

I would use the equation of motion..

V2= U2+2as

where
V2 = 0
U = 60mph = 26.8m/s
s = 0.1m

Solve for "a"

a = -U2/2s
a = - 26.82/(2*0.1)
= -3591m/s2

Thats about 360 "g"

Then..

F=m * a
= 75 * 3591
= 269,325 N

The high force is why cars have crumple zones at the front so you don't stop in 10cm.
 

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