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A car of mass 900kg tows a trailer of mass 600kg by means of a rigid tow bar. The car experiences a resistance of 200N and the trailer a resistance of 300N. If the car engine exerts a forward force of 3kN, find the tension in the tow bar and the acceleration of the system. If the engine is swicthed off and the brakes apply a retarding force of 500N what will be the retardation? Assuming the same resistances? What will be the nature and magnitude of te force of the tow bar?

PLEASE ANSWER!! I've been stuck on it for like two hours now.
Reply 1
Godsize
A car of mass 900kg tows a trailer of mass 600kg by means of a rigid tow bar. The car experiences a resistance of 200N and the trailer a resistance of 300N. If the car engine exerts a forward force of 3kN, find the tension in the tow bar and the acceleration of the system. If the engine is swicthed off and the brakes apply a retarding force of 500N what will be the retardation? Assuming the same resistances? What will be the nature and magnitude of te force of the tow bar?

PLEASE ANSWER!! I've been stuck on it for like two hours now.


I forgot AS maths :frown: I think u use F=MA and you do it for the car and then the trailer. As a result u get 2 simultaneous equations which u shud be able to solve.

3000-200=900-a well wrk out Accelration ne way and then u no wen the drivin force=0 only the resistive ones r left and so u wrk out deceleration and put it into the new formula to wrk out tension.
Reply 2
I've done the first part. The second part I need help with PLEASE!!!
Reply 3
Godsize
A car of mass 900kg tows a trailer of mass 600kg by means of a rigid tow bar. The car experiences a resistance of 200N and the trailer a resistance of 300N. If the car engine exerts a forward force of 3kN, find the tension in the tow bar and the acceleration of the system. If the engine is swicthed off and the brakes apply a retarding force of 500N what will be the retardation? Assuming the same resistances? What will be the nature and magnitude of te force of the tow bar?

PLEASE ANSWER!! I've been stuck on it for like two hours now.


well the force after the engine is turned off is -500 so u get -500-500=1500a so a is -1/3ms^-1 ( i think?!?!?! :confused: ) and then you oh wait is that wrong?
Reply 4
Hey thanks! thats right, Then what do you do?
Reply 5
Net force on system = 3kN - 200N - 300N
= 2,500 N
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Mass of system = 600 + 900 = 1500 kg

accln. of system, a = F/M = 2,500/1,500 = 5/3

a = 5/3 m/s²
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Trailer has accln of 5/3 m/s²

Newton's 2nd law
T- 300 = 600*(5/3) = 1000
T = 1300 N
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Engine switched off
Mass of system = 1500 kg
Net force on system = 500 + 200 + 300 = 1000N, backwards

Retardation produced in the system of 1500 kg mass by a force of 1000N is d, given by

d = F/M = 1000/1500 = 2/3 m/s²
d = 2/3 m/s²
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Tow bar will be in compression with a compressive force C, say, acting against the trailer, and against the direction of motion.

Trailer
force on trailer = 300 + C N, backwards
retardation, d = 2/3 m/s²

d = (300+C)/600
(2/3)*600 = 300 + C
C = 100 N
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