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M2 Moments problem

A uniform ladder of weight W rests at an angle α to the horizontal with its top against a rough vertical wall and its base on rough horizontal ground with coefficient of friction 0.25 at each contact. Find the minimum value of α if the ladder does not slip.

I'm not hitting the correct answer of 61.9 Deg
Could somebody kindly point out the error of my ways?

when taking moments about b, you dont seem to take into account F1
Reply 2
Avatar for w4rtorn
w4rtorn
omg your handwriting and diagram look amazing, mine make people blind and others suicidal.
w4rtorn
omg your handwriting and diagram look amazing, mine make people blind and others suicidal.


that is quite an impressive diagram! i dont think i've ever ever used a ruler (unless i'm doing displacement diagrams where its actually essential)
Reply 4
Avatar for xlaser31
xlaser31
OP
6
didgeridoo12uk
when taking moments about b, you dont seem to take into account F1


The answer should be 61.9 Deg. I can get to 59.5 Deg by taking F1 into account. Any other suggestions?
xlaser31
The answer should be 61.9 Deg. I can get to 59.5 Deg by taking F1 into account. Any other suggestions?


slightly confused as to what r1 and p are in your diagram. (or is it just that you've rubbed r1 out)
Reply 6
Avatar for xlaser31
xlaser31
OP
6
didgeridoo12uk
slightly confused as to what r1 and p are in your diagram. (or is it just that you've rubbed r1 out)

P is the reaction of the ladder against the wall and R1 is the resultant component of that reaction at alpha degrees.
xlaser31
P is the reaction of the ladder against the wall and R1 is the resultant component of that reaction at alpha degrees.


ah ok fair enough. umm i have no idea :/ what you've done looks ok, but i dont have time to work it though myself so...

apart from the usual stuff like check for rounding errors etc...

i'd just flag the question up and ask your teacher next time you see them
Reply 8
Avatar for Noble.
Noble.
17
Using F=uR
F2 = 0.25R2
F1 = 0.25P

Substituting those together and you also get
F1 = 0.25F2

When taking moments, you need to take F1 into account.

wcosx = 2Psinx + 2F1cosx (I'm using x as the angle)

Now, all it is, is rearranging and substituting in values that'll give you it all in terms of F1/F2/P/R2 etc.

R2cosx + F1cosx = 2Psinx + 2F1cosx
R2cosx = 2Psinx + F1cosx
R2cosx = 2F2sinx + F1cosx
R2cosx = 0.5R2sinx + F1cosx
R2cosx = 0.5R2sinx + 0.0625R2cosx

You now have everything in terms of R2, to make it look neater on TSR, I've just multiplied everything through by 16.

16R2cosx = 8R2sinx + R2cosx
15R2cosx = 8R2sinx

Then you'll end up with...

tanx = 15/8 ----> x = 61.928'

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