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M2 Statics of rigid bodies

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Reply 20
Avatar for Kyalimers
Kyalimers
OP
21
OMFG I get this: xmgcos⁡α−2xF2cos⁡α=2xXsin⁡αxmg \cos \alpha - 2xF_2 \cos \alpha = 2xX \sin \alpha

mg+F2=2Xtan⁡αmg+F_2 = 2X \tan \alpha

Y=2Xtan⁡αY=2X \tan \alpha

so 53X=2Xtan⁡α\frac{5}{3}X = 2X \tan \alpha

therefore tan⁡α=56 \tan \alpha =\frac{5}{6} WHICH IS NOT THE tan⁡α=23\tan \alpha = \frac{2}{3} REQUIRED
Reply 21
Avatar for Kyalimers
Kyalimers
OP
21
sohanshah
OMFG I get this: xmgcos⁡α−2xF2cos⁡α=2xXsin⁡αxmg \cos \alpha - 2xF_2 \cos \alpha = 2xX \sin \alpha

mg+F2=2Xtan⁡αmg+F_2 = 2X \tan \alpha

Y=2Xtan⁡αY=2X \tan \alpha

so 53X=2Xtan⁡α\frac{5}{3}X = 2X \tan \alpha

therefore tan⁡α=56 \tan \alpha =\frac{5}{6} WHICH IS NOT THE tan⁡α=23\tan \alpha = \frac{2}{3} REQUIRED

Bump.... anyone tell me where I'm going wrong??
Here's the question again too:

sohanshah
A uniform ladder rests with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ground and the ladder is 35\frac{3}{5} and the coefficient of friction between the wall and the ladder is 13\frac{1}{3}. The ladder is on the point of slipping when it makes an angle α\alpha with the horizontal. Find tan⁡α\tan \alpha.
Reply 22
Avatar for qgujxj39
qgujxj39
17
I did it, and I got tan(alpha) = 5/9, which is different from yours and the required one.
Reply 23
Avatar for Kyalimers
Kyalimers
OP
21
tommmmmmmmmm
I did it, and I got tan(alpha) = 5/9, which is different from yours and the required one.

:s-smilie: Disconcerting.

Weird question huh.
What working did you do?

I call for Bruceleej!!
Reply 24
Avatar for Kyalimers
Kyalimers
OP
21
I still have no idea. Does anyone else. I think tommmmmmmmm is a bit stuck too :s-smilie:
Reply 25
Avatar for bruceleej
bruceleej
2
sohanshah
I still have no idea. Does anyone else. I think tommmmmmmmm is a bit stuck too :s-smilie:

Hi again Sohan, havent been on TSR{maths} for a couple of days. Anyway, Ive managed to solve it. Just need to type it up now...
Reply 26
Avatar for bruceleej
bruceleej
2
Firstly, Resolve Vertically{call this eqn 1}

Spoiler



Secondly, Resolve Horizontally{call this eqn 2};

Spoiler



Now, Take moments about B{eqn 3};

Spoiler



Now Eliminate P and Q;

Spoiler



I hope i havent made any silly errors.
M2.JPG24.7KB
Reply 27
Avatar for Kyalimers
Kyalimers
OP
21
bruceleej
Firstly, Resolve Vertically{call this eqn 1}

Spoiler



Secondly, Resolve Horizontally{call this eqn 2};

Spoiler



Now, Take moments about B{eqn 3};

Spoiler



Now Eliminate P and Q;

Spoiler



I hope i havent made any silly errors.

Lol - I actually did manage to do it in the end but a different way :s-smilie:
Might have fluked it

Thanks (inumerable times) :sugar:

-Sohan-
Reply 28
Avatar for bruceleej
bruceleej
2
No problem, just didnt see it ontime. My browser wouldnt display TSR correctly, its not displaying most images, the TSR logo, the reply buttons etc...

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