Question

In Fig. 12-16, a rigid beam is attached to two posts that are fastened to a floor. A small but heavy safe is placed at the six positions indicated, in turn. Assume that the mass of the beam is negligible compared to that of the safe.

(a) Rank the positions according to the force on post Adue to the safe, greatest compression first, greatest tension last, and indicate where, if anywhere, the force is zero.

(b) Rank them according to the force on post B.

Short answer

(a) The ranking of positions of the 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.

(b) The ranking of positions of the 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.

Step-by-step solution

The given data

The figure for the safe-beam system is given.

 Step 2: Understanding the concept of the force

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, ${\tau }_{net}=0$ (iii)

The torque acting on a point, $\tau =r\times F$ (iv)

a) Calculation of the rank of the safe positions according to the force on post-A

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.

b) Calculation of the rank of the safe positions according to the force on post B

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.