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# Please spot the error. watch

1. I must have made an error somewhere, but i can't quite see where it is.
This is Q27 of Rev ex 2 of the Heinemann M2 textbook.

My working is:
Let S = the magnitude of the force, as the cylinder is smooth the force acts perpendicular to the rod at the point of contact.
Let x be the distance along the rod from A to C.
Let theta=t for simplicity.
Let F be the frictional force and R the contact at the ground.

Resolving upwards:
R + Scos2t = mg

Resolving horizontally:
Ssin2t = F

Taking moments at A gives:
C = AC
(Scos2t)(xcos2t) + (Ssin2t)(xsin2t) = (mg)(3acos2t)
Sx(cos^2 2t + sin^2 2t) = 3mgacos2t
Sx = 3mgacos2t.

Resolving dimensionally upwards:
xsin2t = 2a (diameter of cylinder).
x = 2a/sin2t

Substituting gives:
2aS/sin2t = 3mgacos2t
S = (3mg/2)(cos2t)(sin2t) = (3mg/4)(sin4t).
But the book gives 3mgcos2ttant.

Any help is much appreciated.
2. Gah... wish I could help, but I haven't got that far yet. M2 you say? I may be screwed. Still haven't finished FP1 let alone started M2 and M3.
3. Heh. No worries. You'll get them done ^^
I have P2/P3/P4 to revise and P5/P6/D2 to cover from scratch before June, as well as Bio exams.
4. [QUOTE=Gaz031]

Resolving dimensionally upwards:
xsin2t = 2a (diameter of cylinder).

[QUOTE]

I don't see how you got this bit!

I can't see a diameter in this triangle.

Aitch
5. I have FP1 and P3 to finish and M2 M3 and S2 to do. Did the P2 exam yesterday.
6. [QUOTE=Aitch]
(Original post by Gaz031)

Resolving dimensionally upwards:
xsin2t = 2a (diameter of cylinder).

QUOTE]

I don't see how you got this bit!

I can't see a diameter in this triangle.

Aitch
Put a/x = tan t instead and you've got your answer...

put a = x tant into Sx = 3amg cos2t

get Sx =3x tant mg cos2t =>

3 tant mg cos2t

Aitch
7. [QUOTE=Aitch]
(Original post by Gaz031)

Resolving dimensionally upwards:
xsin2t = 2a (diameter of cylinder).

QUOTE]

I don't see how you got this bit!

I can't see a diameter in this triangle.

Aitch
You mean cylinder?
It's 2.(the radius) = 2a, as the rod rests on the top of the cylinder (2a above the ground).
8. (Original post by Gaz031)
You mean cylinder?
It's 2.(the radius) = 2a, as the rod rests on the top of the cylinder (2a above the ground).
It's not on top of the cylinder - If it were, it couldn't also touch the ground!

Aitch
9. (Original post by Aitch)
Put a/x = tan t instead and you've got your answer...

put a = x tant into Sx = 3amg cos2t

get Sx =3x tant mg cos2t =>

3 tant mg cos2t

Aitch

Aitch
Ahh, i see. Thankyou. The key being that at half the angle it will have only half the elevation.

It's not on top of the cylinder - If it were, it couldn't also touch the ground!
Oh oh my god i'm so stupid

Thanks for the help.
10. (Original post by Gaz031)

Thanks for the help.
No problem! I still owe you for all the exam questions!

Aitch

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