Newton's Second Law Tension in rope Parachute Question✈️
The question below states;
An assortment of military equipment is landed by a parachute, as illustrated in the attached diagram.
The parachute is of mass 20kg and is subject to air resistance of 700N.
Box A has a mass of 60kg and is subject to air resistance of 250N.
Box B is subject to air resistance of 300 N. The ropes between the boxes and the parachute are light and inextensible. given that the parachute and boxes are moving at constant speed, find the tensions in the two ropes and the mass of box B. Assume g=10ms^-2
I think that I need to find the acceleration which ordinarily I would do by using newton's second law;
a=Fnet/mtotal
But I do not know how to do so without the mass of B, nor how to find the mass of B without a?
So to find the tension I initially wrote separate equations of motion for box A and B;
T1 is the tension in the wire linking A to the parachute. T2 is the tension in the wire linking A to B.
A: T2+600N-T1-250N=60a
B: (m*10) -T2-300N=ma
Then to find the tension in both wires solve the pair of simultaneous equations by adding them together. I realise that I have not considered the mass of the parachute but I am not sure where this fits in, would it be in A:
T2+600N-T1-250N-20kg=60a
I am really confused and would greatly appreciate any help/explanation/diagrams to understand this better 👍 Thank you 👌
An assortment of military equipment is landed by a parachute, as illustrated in the attached diagram.
The parachute is of mass 20kg and is subject to air resistance of 700N.
Box A has a mass of 60kg and is subject to air resistance of 250N.
Box B is subject to air resistance of 300 N. The ropes between the boxes and the parachute are light and inextensible. given that the parachute and boxes are moving at constant speed, find the tensions in the two ropes and the mass of box B. Assume g=10ms^-2
I think that I need to find the acceleration which ordinarily I would do by using newton's second law;
a=Fnet/mtotal
But I do not know how to do so without the mass of B, nor how to find the mass of B without a?
So to find the tension I initially wrote separate equations of motion for box A and B;
T1 is the tension in the wire linking A to the parachute. T2 is the tension in the wire linking A to B.
A: T2+600N-T1-250N=60a
B: (m*10) -T2-300N=ma
Then to find the tension in both wires solve the pair of simultaneous equations by adding them together. I realise that I have not considered the mass of the parachute but I am not sure where this fits in, would it be in A:
T2+600N-T1-250N-20kg=60a
I am really confused and would greatly appreciate any help/explanation/diagrams to understand this better 👍 Thank you 👌
To start you off: there is no acceleration.
The question says
'given that the parachute and boxes are moving at constant speed'
So when considering the equation of motion you use, you need to remember that for an object or objects moving at constant speed, there is zero resultant force.
To start a problem like this, it's a good idea to draw a diagram showing all the forces acting on the objects. And then write an equation or two such that you have balanced the forces to give zero resultant.
The question says
'given that the parachute and boxes are moving at constant speed'
So when considering the equation of motion you use, you need to remember that for an object or objects moving at constant speed, there is zero resultant force.
To start a problem like this, it's a good idea to draw a diagram showing all the forces acting on the objects. And then write an equation or two such that you have balanced the forces to give zero resultant.
Thank you for your reply, I have tried to draw a diagram which I have attached here. (sorry ignore the scribbling)
I have derived equations for the parachute and boxes a and b also;
P; 196.2N+T1-700N=20a
A; 588.6N+T2-T1-250N=60a
B;Fmg-T2-300N=ma
I have converted the mass to weight since this question concerns forces. I do not know how to demonstrate that there is zero resultant force here; would it be
196.2N+T1+588.6N+T2+Fmg-700N-250N-T2-300N-20a-60a-ma=0
this appears absurd and i certainly have made a mistake?
I have derived equations for the parachute and boxes a and b also;
P; 196.2N+T1-700N=20a
A; 588.6N+T2-T1-250N=60a
B;Fmg-T2-300N=ma
I have converted the mass to weight since this question concerns forces. I do not know how to demonstrate that there is zero resultant force here; would it be
196.2N+T1+588.6N+T2+Fmg-700N-250N-T2-300N-20a-60a-ma=0
this appears absurd and i certainly have made a mistake?
Your diagrams look ok.
Take g as 10ms-2 as in the question. It makes life easier!
Write down an equation for each object, with the forces upwards balancing the forces downwards.
You will get 3 equations with T1, T2 and m, the mass of the lowest object, as unknowns.
So for the upper object you have the forces there
700N upwards (air resistance)
20g (= 200N) downwards (weight of parachute)
T1 downwards (tension in the rope)
And the equation for zero resultant force means the forces are balanced and there is no acceleration, so
T1 + 200 = 700 (downwards forces = upwards force)
Now you do the same for the other 2 objects.
You then have 3 simultaneous equations to solve.
There is a 4th equation you can use which finds the mass of the lower object directly. If you can't spot that, I'll tell you in my next reply.
Take g as 10ms-2 as in the question. It makes life easier!
Write down an equation for each object, with the forces upwards balancing the forces downwards.
You will get 3 equations with T1, T2 and m, the mass of the lowest object, as unknowns.
So for the upper object you have the forces there
700N upwards (air resistance)
20g (= 200N) downwards (weight of parachute)
T1 downwards (tension in the rope)
And the equation for zero resultant force means the forces are balanced and there is no acceleration, so
T1 + 200 = 700 (downwards forces = upwards force)
Now you do the same for the other 2 objects.
You then have 3 simultaneous equations to solve.
There is a 4th equation you can use which finds the mass of the lower object directly. If you can't spot that, I'll tell you in my next reply.
Original post by Stonebridge
Your diagrams look ok.
Take g as 10ms-2 as in the question. It makes life easier!
Write down an equation for each object, with the forces upwards balancing the forces downwards.
You will get 3 equations with T1, T2 and m, the mass of the lowest object, as unknowns.
So for the upper object you have the forces there
700N upwards (air resistance)
20g (= 200N) downwards (weight of parachute)
T1 downwards (tension in the rope)
And the equation for zero resultant force means the forces are balanced and there is no acceleration, so
T1 + 200 = 700 (downwards forces = upwards force)
Now you do the same for the other 2 objects.
You then have 3 simultaneous equations to solve.
There is a 4th equation you can use which finds the mass of the lower object directly. If you can't spot that, I'll tell you in my next reply.
Take g as 10ms-2 as in the question. It makes life easier!
Write down an equation for each object, with the forces upwards balancing the forces downwards.
You will get 3 equations with T1, T2 and m, the mass of the lowest object, as unknowns.
So for the upper object you have the forces there
700N upwards (air resistance)
20g (= 200N) downwards (weight of parachute)
T1 downwards (tension in the rope)
And the equation for zero resultant force means the forces are balanced and there is no acceleration, so
T1 + 200 = 700 (downwards forces = upwards force)
Now you do the same for the other 2 objects.
You then have 3 simultaneous equations to solve.
There is a 4th equation you can use which finds the mass of the lower object directly. If you can't spot that, I'll tell you in my next reply.
Thank you for your reply,
Parachute: T1+200=700
A: 600 N +T2=T1+250N
B: mgN=300N+T2
Rearranging the first equation;
T1=500N
Substituting into A:
600 N +T2=T1+250N
600 N +T2=500N+250N
600N+T2=750N
T2=150N
Thus, the mass of box b can be found by substituting T2 into B:
mgN=300N+T2
mgN=300N+150N
mgN=450N
m=W/g
m=450/10
m=45kg
Are you saying then to solve
T1+200=700
600 N +T2=T1+250N
mgN=300N+T2
as simulataneous equations ? Or is what I have done enough sufficient instead? 😁 What is the 4th equation to find mass directly ?
You have done it perfectly. All answers correct.
I say 'simultaneous equations' in the general sense that there are 3 equations and 3 unknowns, which means it's possible to find all the unknowns.
In this case they are not so tricky and the simplest method is just as you have used, by substitution.
Yes there is a 4th equation you can use that gets m directly.
You can just look at the 'external' forces on the whole group of objects taken as one system.
So we have:
Upwards:
700N (parachute)
250N (middle object)
300N (lower object)
Downwards
200N (weight of chute)
600N (weight of middle object)
mg (weight of lower object)
The tension forces can be ignored because they are 'internal', and anyway, cancel each other out.
So you could also use
700 + 250 + 300 = 200 + 600 + mg
This will give the same value for m, of course.
Well done.
I say 'simultaneous equations' in the general sense that there are 3 equations and 3 unknowns, which means it's possible to find all the unknowns.
In this case they are not so tricky and the simplest method is just as you have used, by substitution.
Yes there is a 4th equation you can use that gets m directly.
You can just look at the 'external' forces on the whole group of objects taken as one system.
So we have:
Upwards:
700N (parachute)
250N (middle object)
300N (lower object)
Downwards
200N (weight of chute)
600N (weight of middle object)
mg (weight of lower object)
The tension forces can be ignored because they are 'internal', and anyway, cancel each other out.
So you could also use
700 + 250 + 300 = 200 + 600 + mg
This will give the same value for m, of course.
Well done.
Original post by Stonebridge
You have done it perfectly. All answers correct.
I say 'simultaneous equations' in the general sense that there are 3 equations and 3 unknowns, which means it's possible to find all the unknowns.
In this case they are not so tricky and the simplest method is just as you have used, by substitution.
Yes there is a 4th equation you can use that gets m directly.
You can just look at the 'external' forces on the whole group of objects taken as one system.
So we have:
Upwards:
700N (parachute)
250N (middle object)
300N (lower object)
Downwards
200N (weight of chute)
600N (weight of middle object)
mg (weight of lower object)
The tension forces can be ignored because they are 'internal', and anyway, cancel each other out.
So you could also use
700 + 250 + 300 = 200 + 600 + mg
This will give the same value for m, of course.
Well done.
I say 'simultaneous equations' in the general sense that there are 3 equations and 3 unknowns, which means it's possible to find all the unknowns.
In this case they are not so tricky and the simplest method is just as you have used, by substitution.
Yes there is a 4th equation you can use that gets m directly.
You can just look at the 'external' forces on the whole group of objects taken as one system.
So we have:
Upwards:
700N (parachute)
250N (middle object)
300N (lower object)
Downwards
200N (weight of chute)
600N (weight of middle object)
mg (weight of lower object)
The tension forces can be ignored because they are 'internal', and anyway, cancel each other out.
So you could also use
700 + 250 + 300 = 200 + 600 + mg
This will give the same value for m, of course.
Well done.
Thank you for your reply and for showing how to arrive at the fourth equation. Have I really solved it? It seems so much simpler looking at it from this perspective I definitely understand the question to a greater extent and do not feel so perplexed! Thank you very much again 👍😊You are incredibly helpful 👌
You're very welcome.
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