MHB Mechanics- connected particles

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A car tows a caravan down a hill with a slope angle where sin(theta) = 0.05. The car's mass is 1800 kg, and the caravan's mass is 600 kg, with resistive forces of 20N on the car and 80N on the caravan. The thrust force in the tow bar is 50N. Using Newton's laws, the net force equations for both the car and caravan are established, leading to the calculation of the braking force from the car's engine, which is determined to be -420N. The discussion emphasizes the importance of setting equal accelerations for both vehicles to solve for the braking force.
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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.
 
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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
 
skeeter said:
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
 
for the car …

$$F_{net} = 900+50-F_B-20$$

$$a = \dfrac{F_{net}}{M} = \dfrac{930-F_B}{1800} $$

for the caravan …

$$F_{net} = 300 -80-50$$

$$a = \dfrac{F_{net}}{m} = \dfrac{170}{600} = \dfrac{17}{60}[/math]

set the accelerations equal & solve for $F_B$
 
skeeter said:
for the car …

$$F_{net} = 900+50-F_B-20$$

$$a = \dfrac{F_{net}}{M} = \dfrac{930-F_B}{1800} $$

for the caravan …

$$F_{net} = 300 -80-50$$

$$a = \dfrac{F_{net}}{m} = \dfrac{170}{600} = \dfrac{17}{60}[/math]

set the accelerations equal & solve for $F_B$
Thank you so so much!
 
skeeter said:
for the car …

$$F_{net} = 900+50-F_B-20$$

$$a = \dfrac{F_{net}}{M} = \dfrac{930-F_B}{1800} $$

for the caravan …

$$F_{net} = 300 -80-50$$

$$a = \dfrac{F_{net}}{m} = \dfrac{170}{600} = \dfrac{17}{60}[/math]

set the accelerations equal & solve for $F_B$
Where did you get the 900?
 
Aswad Shafin Khan said:
Where did you get the 900?

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