Can anyone have a look at question 5 on this paper I am really stuck.
i can do 5a but i am stuck on 5b
thanks a lot,
btw i have tried but don't seem to get the right answer,
There are 3 forces acting on the beam.
Weight, tension at A, tension at C. You should have gotten
C = 140
A = 70
The beam is in equilibrium, which means no movement. No moments about any point.
No moments about point B!
Let distance CB = x
Moment about B, from force at C:
$\displaystyle M_c = 140(x) $
Moment about B from force at A:
$\displaystyle M_a = 70(x+90) $
Moment from weight(which acts at the centre):
$\displaystyle M_w = 210(\frac{x+90}{2}) $
The sum of these should be zero, and then you have an equation in x to solve!
Remember that anticlockwise moments are positive, and clockwise moments are negative!
The moments from the forces at A and C cause CLOCKWISE motion, which is taken as negative. The weight causes an ANTICLOCKWISE moment, which is positive.
Also you made an error when you went:
$\displaystyle 210(\frac{90+x}{2}) = 210(45+x) $
Fraction do not work like that, it should be:
$\displaystyle 210(\frac{90+x}{2}) = 210(45+\frac{x}{2}) $
How's this look:
$\displaystyle -140(x)-70(x+90)+210(\frac{x+90}{2}) = 0$
$\displaystyle -140(x)-70x-70(90)+210(\frac{x}{2})+210(\frac{90}{2}) = 0$
$\displaystyle -140(x)-70x-70(90)+105x+105(90) = 0$
$\displaystyle -105x = 70(90)-105(90)$
$\displaystyle 105x = 3150$
$\displaystyle x = 30cm$