How Can David Estimate the Weight of a Non-Uniform Log?

[turnstile]

Hi..
Im back :rofl:

A large log AB is 6 m long. It rests in a horizontal position on two smooth supports C and D, where AC = 1 m and BD = 1 m, as shown in Figure . David needs an estimate of the weight of the log, but the log is too heavy to lift off both supports. When David applies a force of magnitude 1500 N vertically upwards to the log at A, the log is about to tilt about D.
(a) State the value of the reaction on the log at C for this case.
... 0...
David initially models the log as uniform rod. Using this model,

(b) estimate the weight of the log
[ I got 3750 N for this]

The shape of the log convinces David that his initial modelling assumption is too simple. He removes the force at A and applies a force acting vertically upwards at B. He finds that the log is about to tilt about C when this force has magnitude 1000 N. David now models the log as a non-uniform rod, with the distance of the centre of mass of the log from C as x metres. Using this model, find

(c) a new estimate for the weight of the log,

im absolutely stuck here... the way I've laid it out is;
Moments at D => 1500x5 =W(4-x)
Moments at C => 1000x5 =?
then... ?

I don't know what to do...
(d) the value of x.

Any help would be appreciated. :)

 

[haynewp]

You basically have it.

Your equation 1500*5-W(4-x)=0 is right
and the other is just 1000*5-W*x=0 (if x is measured to the right of C)

Now you need to solve the 2 equations and 2 unknowns.

 

[turnstile]

I am not getting the answer...

 

[Hootenanny]

I am not getting the answer...

What answer are you getting? And can you show your working please...

~H