Question

The automobile has a weight of 2700 lb and is traveling forward at 4 ft>s when it crashes into the wall. If the impact occurs in 0.06 s, determine the average impulsive force acting on the car. Assume the brakes are not applied. If the coefficient of kinetic friction between the wheels and the pavement is μk = 0.3, calculate the impulsive force on the wall if the brakes were applied during the crash.The brakes are applied to all four wheels so that all the wheels slip.

Answer 1

Answer:

$F_b=153918\, lb.ft.s^{-2}$

Explanation:

Given:

mass, $m=2700 \,lb$

time, $t=0.06\,s$

velocity, $v=4\,ft.s^{-1}$

coefficient of kinetic friction between wheels & pavement, $\mu=0.3$

According to first condition,

$F\times \Delta t=m\times v$

$F\times 0.06= 2700\times 4$

$F=180000\,lb.ft.s^{-2}$

According to second condition,

Magnitude of frictional force (which acts opposite to the direction of motion):

$f=\mu.N$

where N is the normal reaction.

$f=0.3\times 2700\times 32.2$

$f=26082 \,lb.ft.s^{-2}$

Now, the impulsive force on the wall if the brakes were applied during the crash:

$F_b= F-f$

$F_b=180000-26082$

$F_b=153918\, lb.ft.s^{-2}$