Question

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Answer 1

Answer:

(a) 218.6 N

(b) 97.14 N

Step-by-step explanation:

When the system is in equilibrium, the net torque on the system is zero.

AC = 1.5 m, CD = 2.3 m, DB = 5 - 1.5 - 2.3 = 1.2 m

Let the centre of gravity of plank is at G.

AG = 2.5 m, CG = 2.5 - 1.5 = 1 m, GB = 2.5 m

(a) Let the reaction at C is R and at D is R'.

R + R' = 29 x 9.8 = 284.2 N ... (1)

Take the torque about C.

29 x 9.8 x CG = R' x GD

29 x 9.8 x 1 = R' x 1.3

R' = 218.6 N

(b) Take the torque about D.

6 x 9.8 x AD = R x CD

6 x 9.8 x (1.5 + 2.3) = R x 2.3

R = 97.14 N