A uniform beam, XY, 4m long and weighing 350N rests on two pivots P and Q. It is kept in equilibrium by weights of 80N attached at X and 1000N attached at a point between P and Q such that it is 0.6m from Q. If XP = 0.8m and PQ = 2.2m.
(a) calculate the reactions at P and Q ;
(b) if the 1000N weight is replaced with a 1200N weight, at what point from Q should it be placed in order to maintain the equilibrium.
Explanation
(a)
Let A and B be the reactions at P and Q respectively.
A + B = 80 + 350 + 1000 = 1430N
We take moments about Q.
Clockwise moments = 2.2A
Anticlockwise moments = (80 times 3 + 350 times 1 + 1000 times 0.6)
= (1190N)
Both moments are equal
(therefore 2.2A = 1190 implies A = frac{1190}{2.2} = 540.9N)
(B = 1190N – 540.9N = 649.1N)
The reactions at P and Q are 540.9N and 649.1N respectively.
(b)
Reaction at P : ((80 + 350 + 1000)N = 540.9N)
(therefore text{The new reaction at P} = (80 + 350 + 1200)N = frac{1630}{1430} times 540.9)
= (616.6N)
Take moments at Q :
(2.2 times 616.6 = 80 times 3 + 350 times 1 + 1200x)
(1200x = 1356.5 – 590)
(1200x = 766.5)
(x = frac{766.5}{1200})
= (0.63875 approxeq 0.64m)
The 1200N should be placed 0.64m from Q.