Edexcel January 2004 Question 2

A lever consists of a uniform steel rod AB, of weight 100N and length 2m, which rests on a small smooth pivot at a point C of the rod. A load of weight 2200N is suspended from the end B of the rod by a rope. The lever is held in equalibrium in a horizontal position by a vertical force applied at the end A. the rope is modelled as a light string.
Given that BC=0.2m
a) find the magnitude of the force applied at A.

I tried taking moments from C but it just was kinda tragic, so has anyone got the mark scheme or can tell me the answer?
thnx
xxx

Reply 1

mfb3000

At B you have 2200*0.2=440
Weight of the rod causes 100*0.8=80 in the middle
Therefore A has to be equal to 440-80=360. 360/1.8=200N
I know theres no terminology there but thats cause I don't know the proper words to use, if I tried it would just confuse you. Thats why I've just left it as numbers! We haven't got very far in Mechanics yet which is why I have no clue about terminology.