91Ó°ÊÓ

The figure for the safe-beam system is given.

Using the condition for equilibrium, we can write two equations for force and torque for different positions of the safe. Solving these two equations, we can get the values of the magnitudes of the forces at A and B.

Formulae:

The force, according to Newton’s second law,$F_{net}=ma$ (i)

The value of the net force at equilibrium, $F_{net}=0$ (ii)

The value of the torque at equilibrium, ${\mathrm{Ï„}}_{net}=0$ (iii)

The torque acting on a point, $\mathrm{τ}=r×F$ (iv)

Let the weight of the safe is W, and the distance between two positions of the safe on the beam is L.

If the safe is placed at position 1, the force equation at the equilibrium using equation (ii) can be given as:

$F_{\mathit{A}}\mathit{+}{F}_{\mathit{B}}\mathit{=}W\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}a\mathit{\right)}$

From the equilibrium condition of equation (iii), the torque about point B can be given using equation (iv) as:

$$\begin{gathered} F_A\left(2L\right)=W\left(2L\right) \\ {F}_A=W \end{gathered}$$

Using this value in equation (a), we can get the force at B as:

$F_B=0$

If the safe is placed at position 2, the force equation at the equilibrium using equation (ii) can be given as:

$F_A+F_B=W..................\left(b\right)$

From the equilibrium condition of equation (iii), the torque about point B can be given using equation (iv) as:

$$\begin{gathered} F_A\left(2L\right)=W\left(L\right) \\ {F}_A=\frac{W}{2} \end{gathered}$$

Using this value in equation (b), we can get the force at B as:

$F_B=\frac{W}{2}$

If the safe is placed at position 3, the force equation at the equilibrium using equation (ii) can be given as:

$F_A+F_B=W...................\left(c\right)$

From the equilibrium condition of equation (iii), the torque about point B can be given using equation (iv) as:

$$\begin{gathered} F_A\left(2L\right)=W\left(0\right) \\ {F}_A=0 \end{gathered}$$

Using this value in equation (c), we can get the force at B as:

$F_B=W$

If the safe is placed at position 4, the force equation at the equilibrium using equation (ii) can be given as:

$F_A+F_B=W...................\left(d\right)$

From the equilibrium condition of equation (iii), the torque about point B can be given using equation (iv) as:

$$\begin{gathered} F_A\left(2L\right)=-W\left(L\right) \\ {F}_A=-\frac{W}{2} \end{gathered}$$

Using this value in equation (d), we can get the force at B as:

$F_B=\frac{3W}{2}$

If the safe is placed at position 5, the force equation at the equilibrium using equation (ii) can be given as:

$F_A+F_B=W...................\left(e\right)$

From the equilibrium condition of equation (iii), the torque about point B can be given using equation (iv) as:

$$\begin{gathered} F_A\left(2L\right)=-W\left(2L\right) \\ {F}_A=-W \end{gathered}$$

Using this value in equation (e), we can get the force at B as:

$F_B=2W$

If the safe is placed at position 6, the force equation at the equilibrium using equation (ii) can be given as:

$F_A+F_B=W................\left(f\right)$

From the equilibrium condition of equation (iii), the torque about point B can be given using equation (iv) as:

$$\begin{gathered} F_A\left(2L\right)=-W\left(3L\right) \\ {F}_A=-\frac{3}{2}W \end{gathered}$$

Using this value in equation (e), we can get the force at B as:

$F_B=\frac{5W}{2}$

Therefore, the ranking of positions of safe according to the force on A due to the safe is,

$1>2>3>4>5>6$

The force on A is zero at position 3.

From calculations of part (a), the ranking of positions of safe according to the force on B due to the safe is,

$6>5>4>3>2>1$

The force on A is zero at position 1.