A shelf of uniform density is supported by two brackets at a distance of 1/8 and 1/4 of the total length, L, from each end respectively. Find the ratio of the reaction forces from the brackets on the shelf?

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1 Answer
Nov 12, 2017

Translation Equilibrium Condition: F_1+F_2=Mg ........... (1)
Rotational Equilibrium Condition : F_1+6F_2=4Mg ...... (2)

F_1=2/5Mg; \quad F_2=3/5Mg; \qquad F_1/F_2 = 2/3

Explanation:

Translational Equilibrium Condition: \sum_k vec F_k = vec 0
vec F_1 + vec F_2 + Mvec g = 0 \qquad \rightarrow F_1 + F_2 -Mg = 0

F_1 + F_2 = Mg ........ (1)

Rotational Equilibrium Condition: \sum_k vec \tau_k = vec 0
vec \tau_1 + vec \tau_2 + vec \tau_w = 0

vec \tau_1 : Torque due to the force vec F_1
vec \tau_2 : Torque due to the force vec F_2
vec \tau_w : Torque due to the weight of the bar vec w

Calculating the torques about the left end,

vec \tau_1 = +F_1.L/8; \qquad vec \tau_2 = +F_2.(L-L/4)=3/4F_2.L
vec \tau_w = -Mg.L/2

1/8F_1.L + 6/8F_2.L-Mg.L/2 = 0

F_1 + 6 F_2 = 4Mg ....... (2)

We have two equations [(1) and (2)] and two unknowns [F_1 and F_2]. Solving for F_1 and F_2 we get,

F_1 = 2/5Mg; \quad F_2 = 3/5Mg; \qquad F_1/F_2 = 2/3