- Thread starter
- #1

I don't get the ans.

- Thread starter Shah 72
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- Thread starter
- #1

I don't get the ans.

- Mar 1, 2012

- 935

$mg\sin{\theta}$

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

- Thread starter
- #3

Using Newtons law

$mg\sin{\theta}$

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

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

- Mar 1, 2012

- 935

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

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

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 $F_B$

- Thread starter
- #5

Thank you so so much!!!

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

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

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 $F_B$