Question

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?

Answer 1

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`