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M2 moments help

Thought I'd make this thread as I suspect I'll encounter other problems in this chapter/topic.

My Working


The question



Can anyone assist? :smile:
(edited 7 years ago)

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Original post by XxsciencemathsxX
Hi there!

Very unrelated but I was wondering if you could help me with an M2 question because you seem to be great at M2?!


It would probably be best if you made your own thread, Im afraid. M2 is not my strong suit - hence my asking for help.
Original post by XxsciencemathsxX
It's ok!
Just trying to scavenge some help because I really hate M2! :smile:


maybe @Student403 can help?
Reply 3
Hey! Hope this helps. :smile: Let me know if the answer I got matches with that in the textbook.

20160508_170942.jpg
Original post by exo97
Hey! Hope this helps. :smile: Let me know if the answer I got matches with that in the textbook.

20160508_170942.jpg


Thank you. That's very helpful. I thought the force P was applied such that it made an angle 30 degrees to the horizontal but i see now it's the rod.
Reply 5
Original post by Kvothe the arcane
Thank you. That's very helpful. I thought the force P was applied such that it made an angle 30 degrees to the horizontal but i see now it's the rod.


No problem :smile:

Question



Working



Model in paint



Can anyone assist with this?

I took moments about X and as the system is in equilibrium, I've said that 6gcos10+2Tycos55cos80=2Tysin55cos6g\cos10 + 2T_y \cos 55 \cos 80 = 2T_y \sin 55 \cos which leads to Ty=57.91.4=41.4NT_y=\dfrac{57.9}{1.4}=41.4N which is incorrect. As this is incorrect, my other results would also be wrong so I'm stuck.

Is my model wrong?
(edited 7 years ago)
Original post by Kvothe the arcane

Question



Working



Model in paint



Can anyone assist with this?

I took moments about X and as the system is in equilibrium, I've said that 6gcos10+2Tycos55cos80=2Tysin55cos6g\cos10 + 2T_y \cos 55 \cos 80 = 2T_y \sin 55 \cos which leads to Ty=57.91.4=41.4NT_y=\dfrac{57.9}{1.4}=41.4N which is incorrect. As this is incorrect, my other results would also be wrong so I'm stuck.

Is my model wrong?


Should the answer be 40.9? If so, all that is wrong is that you have rounded your numbers off before you got to the final answer.
Original post by tiny hobbit
Should the answer be 40.9? If so, all that is wrong is that you have rounded your numbers off before you got to the final answer.


Thanks for replying :smile:. The answers are Ty=37.4NT_y=37.4N, Tx=31.9NT_x=31.9N and 29.329.3^{\circ} to the vertical.

I asked a friend who approached the problem differently by considering a converging triangle of forces.

Working



I follow it but I'm unsure what it is I did wrong. Can you help? :please:
Original post by Kvothe the arcane
Thanks for replying :smile:. The answers are Ty=37.4NT_y=37.4N, Tx=31.9NT_x=31.9N and 29.329.3^{\circ} to the vertical.

I asked a friend who approached the problem differently by considering a converging triangle of forces.

Working



I follow it but I'm unsure what it is I did wrong. Can you help? :please:


I also did it with a triangle of forces (effectively) and got the same answer as you, give or take the rounding issue.

The difference between your friend's version and ours is the direction the strings are going in. You and I both had them going outwards from the ends of the rod, he has them coming inwards.

A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.
Reply 10
You should also practice resolving distances rather than forces - I find this makes certain scenarios easier to deal with.
Original post by tiny hobbit
I also did it with a triangle of forces (effectively) and got the same answer as you, give or take the rounding issue.

The difference between your friend's version and ours is the direction the strings are going in. You and I both had them going outwards from the ends of the rod, he has them coming inwards.

A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.


I repeated the question drawing the forces coming inwards and I still got the same answer I'm afraid. I honestly don't get what's wrong with what I did (beyond the rounding error). What's the significance of "vertically above the rod" sorry? I'm sorry to be a bit of a nuisance here but it's a bit frustrating not knowing what went wrong. I wouldn't have an answer in the exam obviously.

Original post by Ayman!
You should also practice resolving distances rather than forces - I find this makes certain scenarios easier to deal with.


Thanks. I think that was the point of the exercise :colondollar:.

13115873_634681436689463_2093575599_n.jpg

Do you know why it doesn't work out for if I use 6g as part of the triangle? Like in the example? What's different about the example?

Would you mind assisting @ghostwalker please? When I used the triangle my friend set up, I was able to follow it to get to the correct answer but I'm uncomfortable not knowing what went wrong with my working. Even when I drew my lines going inwards, I got the same figures and answer.
(edited 7 years ago)
Original post by Kvothe the arcane
I
Would you mind assisting @ghostwalker please?


I'm not clear which image you want me to look at.
Original post by ghostwalker
I'm not clear which image you want me to look at.


Hi, there sorry. The original question is reposted below.
I redrew the diagram with the forces pointed inwards and taking moments about X I got 6gcos10+2Tysin35sin10=2Tycos35cos10Ty=57.91.413=40.986g \cos10 + 2T_y \sin 35 \sin 10=2T_y cos 35 \cos 10 \Rightarrow Ty=\dfrac{57.9}{1.413}=40.98. There may be a rounding error but still, it's identical to what I've below and far from the actual answer. I'm unsure why it doesn't work.

Tiny Hobbit has said A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.

But the significance of what she says is lost on me. Does it affect my model?

Thank you

Question


Working


Model in paint

(edited 7 years ago)
Original post by Kvothe the arcane
Hi, there sorry. The original question is reposted below.
I redrew the diagram with the forces pointed inwards and taking moments about X I got 5gcos10+2Tysin35sin10=2Tycos35cos10Ty=57.91.413=40.985g \cos10 + 2T_y \sin 35 \sin 10=2T_y cos 35 \cos 10 \Rightarrow Ty=\dfrac{57.9}{1.413}=40.98. There may be a rounding error but still, it's identical to what I've below and far from the actual answer. I'm unsure why it doesn't work.


Your rod seems to have become lighter. 5kg rather than the 6kg in the question.
Your Tysin35sin10T_y\sin 35\sin10 should be going anticlockwise and as such on the RHS of your equation (with the 2, of course).


Tiny Hobbit has said A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.

But the significance of what she says is lost on me. Does it affect my model


Yes - your model (final image) is of the orignal set up.
(edited 7 years ago)
Original post by ghostwalker
Your rod seems to have become lighter. 5kg rather than the 6kg in the question.
Your Tysin35sin10T_y\sin 35\sin10 should be going anticlockwise and as such on the RHS of your equation (with the 2, of course).

Yes - your model (final image) is of the orignal set up.


Thanks ghostwalker. The 6 was a typo on here and I suppose it makes sense for the split components of the Ty to be both upwards if it was upwards in the first place.

Thanks again for your help. I got 31.95 which is not quite 31.9 (another rounding error) which suggests I should use more d.p in prior steps than I do or keep the trigfunction(value) until the last step.
(edited 7 years ago)
Original post by Kvothe the arcane
I suppose it makes sense for the split components of the Ty to be both upwards if it was upwards in the first place.


That sounds as if you're a bit unsure. The Ty sin 35 is above the line of the rod, so if you want the component perpendicular to the rod, it will be above, hence anticlockwise.



Thanks again for your help. I got 31.95 which is not quite 31.9 (another rounding error) which suggests I should use more d.p in prior steps than I do or keep the trigfunction(value) until the last step.


As a general guide, I delay putting in actual numbers as long as possible - you often find the algebra will simplify.

Regarding the question in general:
If you'd done a bit of elmentary geometry at the start, and resolved Ty parallel and perpenedicular to the rod (rather than going for vertical and horizontal as an intermediate resolution), the algebra would have been a lot simpler.

6gcos10=2Tycos256g\cos 10 = 2T_y\cos 25
Original post by ghostwalker
That sounds as if you're a bit unsure. The Ty sin 35 is above the line of the rod, so if you want the component perpendicular to the rod, it will be above, hence anticlockwise.

Thank :smile:


If you'd done a bit of elmentary geometry at the start, and resolved Ty parallel and perpenedicular to the rod (rather than going for vertical and horizontal as an intermediate resolution), the algebra would have been a lot simpler.

6gcos10=2Tycos256g\cos 10 = 2T_y\cos 25

Oh, of course :facepalm2:. Thanks
I'll type my next question this time as I know that pictures aren't always very clear

But it's question 8 in exercise 5D of the Pearsons Edexcel book if you have it to hand

Question
A uniform ladder rests in limiting equilibrium with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ladder and the ground is μ1\mu_1 and the coefficient of friction between the ladder and the wall is μ2\mu_2. Given the ladder makes and angle θ\theta with the horizontal, show that tanθ=1μ1μ22μ1\tan \theta=\dfrac{1- \mu_1 \mu 2}{2\mu_1 }


R()RA+F2=WR()RB=F1R(\uparrow)R_A+F_2=W \\ R(\rightarrow)R_B=F_1

M:A   xWcosθ=2xRbsinθ+2xF2cosθ M:A \ \ \ xW \cos \theta = 2xR_bsin \theta + 2xF_2 \cos \theta

cosθ(WF2)=2RBsinθW2F22Rb=tanθ=RaF22RB\therefore \cos \theta(W-F_2)=2R_Bsin \theta \Rightarrow \dfrac{W-2F_2}{2R_b}=tan \theta=\dfrac{Ra-F_2}{2R_B}

μ1=F1RAμ2=F2RB\mu_1=\frac{F_1}{R_A} \quad \mu_2=\frac{F_2}{R_B}
tanθ=1μ1μ22μ1=1F1RAF2RB2F1RA=F1(RAF2)2F1Rb=RaF22RB=tanθ\tan \theta=\dfrac{1- \mu_1 \mu 2}{2\mu_1 }=\dfrac{1-\frac{F_1}{R_A} \frac{F_2}{R_B}}{\frac{2F_1}{R_A}} = \dfrac{F_1(R_A-F_2)}{2F_1Rb}= \dfrac{Ra-F2}{2R_B}= \tan \theta as required.
So my question is that if I've shown tanθ=...\tan \theta = ... and I show that something else is =...=..., is that fine? It seemed simpler than changing ...... into the LHS of the question.
(edited 7 years ago)
Original post by Kvothe the arcane
I'll type my next question this time as I know that pictures aren't always very clear

But it's question 8 in exercise 5D of the Pearsons Edexcel book if you have it to hand



R()RA+F2=WR()RB=F1R(\uparrow)R_A+F_2=W \\ R(\rightarrow)R_B=F_1

M:A   xWcosθ=2xRbsinθ+2xF2cosθ M:A \ \ \ xW \cos \theta = 2xR_bsin \theta + 2xF_2 \cos \theta

cosθ(WF2)=2RBsinθW2F22Rb=tanθ=RaF22RB\therefore \cos \theta(W-F_2)=2R_Bsin \theta \Rightarrow \dfrac{W-2F_2}{2R_b}=tan \theta=\dfrac{Ra-F_2}{2R_B}

μ1=F1RAμ2=F2RB\mu_1=\frac{F_1}{R_A} \quad \mu_2=\frac{F_2}{R_B}
tanθ=1μ1μ22μ1=1F1RAF2RB2F1RA=F1(RAF2)2F1Rb=RaF22RB=tanθ\tan \theta=\dfrac{1- \mu_1 \mu 2}{2\mu_1 }=\dfrac{1-\frac{F_1}{R_A} \frac{F_2}{R_B}}{\frac{2F_1}{R_A}} = \dfrac{F_1(R_A-F_2)}{2F_1Rb}= \dfrac{Ra-F2}{2R_B}= \tan \theta as required.
So my question is that if I've shown tanθ=...\tan \theta = ... and I show that something else is =...=..., is that fine? It seemed simpler than changing ...... into the LHS of the question.


I'm not quite sure what you mean. Do you mean leaving it as the tantheta =forces part?

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