# A level Maths help mechanics

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Thread starter 1 week ago
#1
A tractor and its driver have a combined mass of 4m kilograms.
The tractor is towing a trailer of mass m kilograms in a straight line along a horizontal
road.
The tractor and trailer are connected by a horizontal tow bar, modelled as a light rigid
rod.
A driving force of 11 080 N and a total resistance force of 160 N act on the tractor.
A total resistance force of 600 N acts on the trailer.
The tractor and the trailer have an acceleration of 0.8 m s2

Work out m and the tension in the tow bar

At the instant the speed of the tractor reaches 18 km h1 the tow bar breaks.
The total resistance force acting on the trailer remains constant.
Starting from the instant the tow bar breaks, calculate the time taken until the speed
of the trailer reduces to 9 km h1
How would I go about working this out?
Last edited by ayman98763; 1 week ago
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Thread starter 1 week ago
#2
Plsss i need help
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1 week ago
#3
What have you got so far?
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1 week ago
#4
(Original post by ayman98763)
Plsss i need help
Are you just asking us to do your homework for you or do you need help with something specific?
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1 week ago
#5
First, try using F=ma to create equations of motion for the trailer and the tractor.
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Thread starter 1 week ago
#6
(Original post by Nagromicous)
Are you just asking us to do your homework for you or do you need help with something
I need help working out the tension and time taken? I've worked out the mass
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1 week ago
#7
(Original post by ayman98763)
I need help working out the tension and time taken? I've worked out the mass
What is the resultant force, mass and acceleration of the trailer?
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Thread starter 1 week ago
#8
(Original post by squidge_gt)
What have you got so far?
ive got that 11080-600-160=0.8*m
and then i got the mass as 12900Newtons
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Thread starter 1 week ago
#9
(Original post by Zelerate)
What is the resultant force, mass and acceleration of the trailer?
ive got the mass as 12900newtons
acceleration is 0.8ms^-2
and i think resultant force is 11080-600-160=10320 newtons
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1 week ago
#10
If you have the mass of the tractor + trailer (5m) to be 12900, then you can figure out m. Then consider the tractor forces: 11080-160-T=4m * 0.8
Last edited by Zelerate; 1 week ago
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Thread starter 1 week ago
#11
(Original post by Zelerate)
If you have the mass of the tractor + trailer (5m) to be 12900, then you can figure out m. Then consider the tractor forces: 11080-160-T=m * 0.8
so 11080-600-T = 2580*4
10480-T =10320
so tension is 160 Newtons
is that right?
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1 week ago
#12
(Original post by ayman98763)
so 11080-600-T = 2580*4
10480-T =10320
so tension is 160 Newtons
is that right?
Multiply 2580*4 by 0.8.
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Thread starter 1 week ago
#13
(Original post by Zelerate)
Multiply 2580*4 by 0.8.
2580*4=10320*0.8=8256
10480-8256=2224 newtons
so 2224 newtons as tension.
Any idea about working out time taken, would that involve suvat equations?
Last edited by ayman98763; 1 week ago
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1 week ago
#14
(Original post by ayman98763)
2580*4=10320*0.8=8256
so 8256 newtons as tension.
Any idea about working out time taken, would that involve suvat equations?
11080 - 160 - 2580*4*0.8 = T = 2664.

For the time taken, it would indeed involve suvat equations as the acceleration is constant. v = u + at would be useful here. Since you know the force and the mass, you can work out the acceleration.
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Thread starter 1 week ago
#15
(Original post by Zelerate)
11080 - 160 - 2580*4*0.8 = T = 2664.

For the time taken, it would indeed involve suvat equations as the acceleration is constant. v = u + at would be useful here. Since you know the force and the mass, you can work out the acceleration.
so i converted kmh^-1 to ms^-1

s
u = 5ms-1 f=ma 10320/12900=0.8 i just get the same acceleration
v = 3.333ms-1
a =0.8
t ?
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1 week ago
#16
(Original post by ayman98763)
so i converted kmh^-1 to ms^-1

s
u = 5ms-1 f=ma 10320/12900=0.8 i just get the same acceleration
v = 3.333ms-1
a =0.8
t ?
Using F = ma (and a = F/m), the acceleration of the trailer should be -600/2560 as the tension no longer acts on it, and the mass is one fifth of 12900 (5m / 5)

Also, as the speed is going from 18 km/h to 9 km/h, shouldn't your value of v should be 2.5?
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Thread starter 1 week ago
#17
-600/12900/5= -600/2580 =-10/43

v=u + a t
2.5= 5 +(-10/43)t
-2.5/-10/43 =1075ms-1
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Thread starter 1 week ago
#18
(Original post by Zelerate)
Using F = ma (and a = F/m), the acceleration of the trailer should be -600/2560 as the tension no longer acts on it, and the mass is one fifth of 12900 (5m / 5)

Also, as the speed is going from 18 km/h to 9 km/h, shouldn't your value of v should be 2.5?
yeah my bad I read 12 instead of 9
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