A & B is 2 points on a plank on 2 trestles 3m apart. A man standing 1m from A with 700N & a block of concrete 1m from the man with..
400N. If the man walks past A towards the end of the plank,what is the reaction force at B & how far the man will be from A as the plank starts to tip?
We suppose that the plank XY is resting on the trestles at A and B which are 3 meters apart. Further the mass of the plank plank m is not given. The total length of the plank L is also not given. We, obviously, pressume L to be very much greater than 3 meter. Also we pressume that L is symmetrically supported on the trestles at A and B.
Let , M , C and G be the points on the plank where the forces of weight of the man, concrete block and the mass of the plank act on the plank. Let Ra and Rb be the reactions at A and B.
By the given data the picture is now like:
AM =1 MC =1 , given AG = 1.5 meter, as the plank rests symmetrically of A and B.
Since the system is in equilibrium, the net force and the moment are each equal to zero.Therefore,
Ra+Rb+700+400+mg = 0, the net force...................(1)
Ra*0-(AM*700+AC*400+AG*mg+AB*Rb) = 0, the net moment about A. Or,
0-(1*700+2*400+1.5*mg+3*Rb) = 0 ...................................(2), minus sign is due to moments about A is considered clokwise.
From (2) Rb1 = -(1500+1.5mg)/3 N or (500+0.5mg) N upwards.
As the man walks past A, at a distance x past A, the moments about A are:
700*x -(2*400+1.5mg+3*Rb) = 0 or,
3Rb - (700*x-800-1.5mg) = o or,
Rb = (700*x-800-1.5mg)/3 is the reation at B, when the man is x meter away past A. Or
Rb - (700*x-800)/3 if the weight of the plank is to be neglected.
When the planks at B just about to topple, Rb = 0. Or the reaction force at B is zero. So, 0= (700x-800-1.5mg)/3 or,
700x = 800+1.5mg 0r
x = (800+1.5mg)/700 = (8/7 +1.5mg/700) meter past A , beyond which if the man walks, the plank falls.
x= 8/7 meter if the plank weight is negligible. And Rb = 0.