# M1 Connected Particles - Help, please?

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#1
I'm having trouble understanding how to complete this seemingly simple M1 question:

I understand that being light and a rod, it is inextensible with no mass, and I understand that you have to take the entire weight of the system, along with the force acting on it - 15N, to find the acceleration of the entire system, which is 2.2ms^-2. But after this, I find it difficult to visualise what you have to do next. Is there one specific particle out of the two that you have to focus on, or can it be either particle? My thought process is that the total 'up' force experienced by the lower particle, is 15N, plus the tension/thrust T in the string so resultant up is (15+T), with its weight, 0.75kg, acting against this. The resultant force is (15 + T - 0.75g), which is equal to 0.75a [F = MA]. This gives a final value for tension as -6N, and not 6N, the correct answer.

My reasoning for the top particle, is that it has the thrust 15N acting on it in the 'up' direction, and it's own weight, plus the weight of the other mass, and the tension in the string, acting down. So, the resultant force is 15 - 1.25g - T, equal to 0.5a. This gives the tension to be 1.65, which isn't correct at all. What am I missing here? Thanks.
0
2 years ago
#2
(Original post by Dggj_19)
I'm having trouble understanding how to complete this seemingly simple M1 question:

I understand that being light and a rod, it is inextensible with no mass, and I understand that you have to take the entire weight of the system, along with the force acting on it - 15N, to find the acceleration of the entire system, which is 2.2ms^-2. But after this, I find it difficult to visualise what you have to do next. Is there one specific particle out of the two that you have to focus on, or can it be either particle? My thought process is that the total 'up' force experienced by the lower particle, is 15N, plus the tension/thrust T in the string so resultant up is (15+T), with its weight, 0.75kg, acting against this. The resultant force is (15 + T - 0.75g), which is equal to 0.75a [F = MA]. This gives a final value for tension as -6N, and not 6N, the correct answer.
For the 0.75kg mass you assumed that T was acting upwards. Getting a negative value, simply means that it is acting in the opposite direction to the one you assumed - i.e. it's downwards. For the lower mass, a "tension" would act upwards, and a "thrust" would act downwards. The rod is exerting a thrust on the lower mass.

My reasoning for the top particle, is that it has the thrust 15N acting on it in the 'up' direction, and it's own weight, plus the weight of the other mass, and the tension in the string, acting down. So, the resultant force is 15 - 1.25g - T, equal to 0.5a. This gives the tension to be 1.65, which isn't correct at all. What am I missing here? Thanks.
For the top mass, using F=ma, you only consider forces acting directly on the mass (as you did with the lower mass).
The 15N force is not acting on the top mass. Only the weight of the mass, and the thrust in the rod act on that mass.
T will act upwards on the mass.
Its weight will act downards.

Edit: You can work with either particle.
Last edited by ghostwalker; 2 years ago
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#3
(Original post by ghostwalker)
For the 0.75kg mass you assumed that T was acting upwards. Getting a negative value, simply means that it is acting in the opposite direction to the one you assumed - i.e. it's downwards. For the lower mass, a "tension" would act upwards, and a "thrust" would act downwards. The rod is exerting a thrust on the lower mass.

For the top mass, using F=ma, you only consider forces acting directly on the mass (as you did with the lower mass).
The 15N force is not acting on the top mass. Only the weight of the mass, and the thrust in the rod act on that mass.
T will act upwards on the mass.
Its weight will act downards.

Edit: You can work with either particle.
Ah, so, thrust acts in the opposite direction to travel, right? It seems obvious saying it but I was getting tangled up in it all and I was thinking of thrust as just the force forward. Just a few things to clarify, though.

The question states that the force upwards is being exerted at the particle Q. How is the rod exerting any force on the lower particle when the lower particle is pushing the rod? It makes sense that the upwards force causes the rod to act on the top particle, but not the bottom one. I see now that my reasoning for the top particle was incorrect because I was imagining a force pulling the system up from the top particle, hence the weight of the other particle and the tension acting against it. But in the scenario, the system is being pushed from the bottom, and as such, the rod is applying a force on the top particle upwards, so shouldn't the thrust supplied by the rod not be upwards, but downwards?
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2 years ago
#4
(Original post by Dggj_19)
Ah, so, thrust acts in the opposite direction to travel, right?
The thrust acts downwards on Q, but upwards on P, so it acts in both directions depending on where you're considering it.

The question states that the force upwards is being exerted at the particle Q. How is the rod exerting any force on the lower particle when the lower particle is pushing the rod? It makes sense that the upwards force causes the rod to act on the top particle, but not the bottom one.
Newton's 3rd law. To every action, there is an equal and opposite reaction. Q exerts a force on the rod, and the rod exerts a force on Q.

But in the scenario, the system is being pushed from the bottom, and as such, the rod is applying a force on the top particle upwards, so shouldn't the thrust supplied by the rod not be upwards, but downwards?
The thrust is downwards at Q, and upwards at P.

At opposite ends of a rod, thrust works in opposite directions. Also tension works in opposite directions at opposite ends of a rod.
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#5
(Original post by ghostwalker)
The thrust acts downwards on Q, but upwards on P, so it acts in both directions depending on where you're considering it.

Newton's 3rd law. To every action, there is an equal and opposite reaction. Q exerts a force on the rod, and the rod exerts a force on Q.

The thrust is downwards at Q, and upwards at P.

At opposite ends of a rod, thrust works in opposite directions. Also tension works in opposite directions at opposite ends of a rod.
I still don't understand how the thrust can be acting in opposite directions for both particles, when they are both individually being pushed upwards by a force, from the bottom - for the bottom one, by the given 15N force, and for the top, the rod pushing against it - and the particles move in the same direction. The quote ' The thrust acts downwards on Q, but upwards on P ' is very counterintuitive to me - how/why is this, exactly?
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2 years ago
#6
(Original post by Dggj_19)
I still don't understand how the thrust can be acting in opposite directions for both particles, when they are both individually being pushed upwards by a force, from the bottom - for the bottom one, by the given 15N force, and for the top, the rod pushing against it - and the particles move in the same direction. The quote ' The thrust acts downwards on Q, but upwards on P ' is very counterintuitive to me - how/why is this, exactly?
For P.

Acceleration is upwards, and the force is being tranmitted to P via the rod, so the force in the rod at that point must be upwards.

For Q.

If the rod wasn't there, the mass Q would accelerate upwards more quickly, The presence of the rod (with P on top) is reducing that upwards acceleration, so it must be exerting a downwards force on Q.

A different situation.
Take a strong rod/ruler/whatever. Hold one end in your right hand and one in your left. Now try and bring your hands together.
Your right hand can't move to the left because the rod exerts a force acting to the right at that point.
At the other end, your left hand can't more to the right, as the rod is exerting a force acting to the left at that point.

A similar scenario if you try and pull the two ends apart, and get thrust in the rod; forces will be the otherway around there.
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#7
Still a difficult thing to try to picture accurately but I believe I understand better now. Thank you!
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