Turning effects, moments, levers, determing the magnitude of the force

[lolyland]

The figure below shows a section of the human forearm in equilibrium.


The weight of the object in the hand is 60 N. The centre of gravity of this object is
32 cm from the elbow. The bicep provides an upward force of magnitude F. The distance between the line of action of this force and the elbow is 3.5 cm. The weight of the forearm is 18 N. The distance between the centre of gravity of the forearm and the elbow is 14 cm.
By taking moments about the elbow, determine the magnitude of the force F provided by the bicep.

(Diagram Attached)

I have attempted at the question
Ʃ of moment in clockwise = Ʃ of moment in anticlockwise
F1 x S1 + F2 x S2

F3 xS3

However, I am rather lost!
Any help would be great thanks!

 

[Simon Bridge]

Start by writing down the moments about the elbow.

 

[CWatters]

I have attempted at the question
Ʃ of moment in clockwise = Ʃ of moment in anticlockwise

You seem to have the write approach so not sure why you are lost.

Especially if you meant to write..

F1 x S1 + F2 x S2 = F3 x S3

 

[lolyland]

My second attempt!

ƩMc=ƩMA
F1S1 + F2S2 =F3xS3
60 x 0.32 + 18x0.14 = F3 x 0.035
19.2 + 2.52 = 0.035 F3
22.72= 0.035 F3
21.62/0.035= F3
620.6 N = F3

I am hoping this is right!

 

[CWatters]

That's what I make it.