Mechanics- connected particles

[Shah 72]

A car tows a caravan down a hill. The slope of the hill makes an angle theta with the horizontal, where sin theta = 0.05. The force from the car's engine is a braking force ( a negative driving force). The car has mass 1800 kg and the caravan has mass 600kg. The resistance on the car is 20N and that on the caravan is 80N. The force in the tow bar is a thrust of 50N. Show that the force from the car's engine is -420N.
I don't get the ans.

 

[skeeter]

forces acting on the caravan down the incline …
$mg\sin{\theta}$
up the incline …
80N resistive force + 50N thrust force from the tow bar

forces acting on the car down the incline …
$Mg\sin{\theta} +$ 50N thrust force from the tow bar
up the incline …
20N resistive force + $F_B$, the braking force

both the car & caravan have the same acceleration

try again

 

[Shah 72]

forces acting on the caravan down the incline …
$mg\sin{\theta}$
up the incline …
80N resistive force + 50N thrust force from the tow bar

forces acting on the car down the incline …
$Mg\sin{\theta} +$ 50N thrust force from the tow bar
up the incline …
20N resistive force + $F_B$, the braking force

both the car & caravan have the same acceleration

try again

Using Newtons law
F= m×a
18000sintheta+ 6000sintheta-20-50-80=2400a
a=0.438m/s^2
Again using F=m×a
18000×0.05-50-20-braking force= 1800×0.438
I don't get the ans

 

[skeeter]

for the car …

\(\displaystyle F_{net} = 900+50-F_B-20\)

\(\displaystyle a = \dfrac{F_{net}}{M} = \dfrac{930-F_B}{1800} \)

for the caravan …

\(\displaystyle F_{net} = 300 -80-50\)

\(\displaystyle a = \dfrac{F_{net}}{m} = \dfrac{170}{600} = \dfrac{17}{60}\)

set the accelerations equal & solve for $F_B$

 

[Shah 72]

for the car …

\(\displaystyle F_{net} = 900+50-F_B-20\)

\(\displaystyle a = \dfrac{F_{net}}{M} = \dfrac{930-F_B}{1800} \)

for the caravan …

\(\displaystyle F_{net} = 300 -80-50\)

\(\displaystyle a = \dfrac{F_{net}}{m} = \dfrac{170}{600} = \dfrac{17}{60}\)

set the accelerations equal & solve for $F_B$

Thank you so so much!

 

[Aswad Shafin Khan]

for the car …

\(\displaystyle F_{net} = 900+50-F_B-20\)

\(\displaystyle a = \dfrac{F_{net}}{M} = \dfrac{930-F_B}{1800} \)

for the caravan …

\(\displaystyle F_{net} = 300 -80-50\)

\(\displaystyle a = \dfrac{F_{net}}{m} = \dfrac{170}{600} = \dfrac{17}{60}\)

set the accelerations equal & solve for $F_B$

Where did you get the 900?

 

[skeeter]

Where did you get the 900?

$mg\sin{\theta} = 900 \, N$ for $g \approx 10 \, m/s^2$