Question

An 1100 kg trailer is hitched to a 3200 kg car. The car and trailer are traveling at 60 km/h when the driver applies the brakes on both the car and trailer.

Knowing that the braking forces exerted on the car and the trailer are 3700 N and 800 N respectively, determine

(a) the distance traveled [m] by the car and trailer before they come to stop,

(b) the horizontal component of the force [N] exerted by trailer hitch onto the car.

Answer 1

Answer:

(a). The distance traveled by the car and trailer before they come to stop is 132.32 m

(b). The horizontal component of the force [N] exerted by trailer hitch onto the car is 2560 N.

Explanation:

Given that,

Mass of trailer = 1100 kg

Mass of car = 3200 kg

Speed of car and trailer = 60 km/h =16.67 m/s

Braking force on the car= 3700 N

Braking force on the trailer = 800 N

We need to calculate the acceleration

Using formula of acceleration

$F=ma$

$a=\dfrac{F}{m}$

Where, F = total force

m = total mass

Put the value into the formula

$a=\dfrac{3700+800}{1100+3200}$

$a=1.05\ m/s^2$

(a). We need to calculate the distance traveled by the car and trailer before they come to stop

Using equation of motion

$v^2=u^2-2as$

Put the value into the formula

$0=(16.67)^2-2\times1.05\times s$

$s=\dfrac{277.88}{2\times1.05}$

$s=132.32\ m$

(b). We need to calculate the horizontal component of the force exerted by trailer hitch onto the car

Firstly, we need to calculate the average braking force

$F=ma$

Put the value into the formula

$F=3200\times1.05$

$F=3360\ N$

We need to calculate the net force on hitch

$F=3360-800$

$F=2560\ N$

Hence, (a). The distance traveled by the car and trailer before they come to stop is 132.32 m

(b). The horizontal component of the force [N] exerted by trailer hitch onto the car is 2560 N.