Question

A $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{-}\mathit{N}$350-N, uniform, $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{-}\mathit{m}$1.50-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{-}\mathit{N}$ 400.0 N. You want to place a small weight on this bar.

(a) What is the heaviest weight you can put on without breaking either cable, and

(b) where should you put this weight?

Short answer

(a) The heaviest object that can be put on the bar weighs $550.0\mathit{}N$.

(b) The weight should be put at a distance of 0.614 meters from point A.

Step-by-step solution

Step 1: Equilibrium 

The condition for translational equilibrium is:$\underset{}{\mathbf{\sum }{\mathit{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}}$, And that for rotational equilibrium is: $\underset{}{\mathbf{\sum }{\mathbf{\pi }}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}}$. The vector sum of all the forces will be zero.

Find the Weight

(a)

Cable A can support the maximum tension$T_A=500.0N$ , and cable B can support the maximum tension of $T_B=400.0N$.

Let the whole given setup be illustrated as a free body diagram for tension in the cables and the gravitational weight, as shown in the figure as:

Here, the weight of the heaviest object to be put on the bar weighing 350 N is indicated as $\stackrel{\mathit{u}\mathit{u}}{W}$.

Considering the upward force to be positive and applying the condition for translational equilibrium, we have:

$$\begin{gathered} \underset{}{\sum F_y=T_A+T_B-350-W=0} \\ W=T_A+T_B-350 \\ =500+400-350 \\ =550N \end{gathered}$$

Thus, the heaviest object that can be put on the bar weighs $550.0N$.

Step 3: Find the position

(b)

Let, the weight to be put at a distance $xmeter$from point A.

Again, considering the anticlockwise rotation to be positive and applying the condition for rotational equilibrium, we have:

$$\begin{gathered} \underset{}{\sum {\mathrm{\pi }}_{ext}={\mathrm{T}}_{\mathrm{B}}.\left(1.50-\mathrm{x}\right)}-\left(350.0\right).\left(\frac{1.50}{2}-x\right)-T_A.x=0 \\ {T}_B.I-T_B.x+350x-T_A.x=\left(350.0\right).\left(\frac{1.50}{2}\right) \\ x=\frac{\left(400.0\right).\left(1.50\right)-\left(350.0\right).\left({\displaystyle \frac{1.50}{2}}\right)}{400.0-350+500} \\ =0.614m \end{gathered}$$

Thus, the weight should be put at a distance of 0.614 meters from point A.