Q9 E - University Physics with Modern Physics

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?

 

Step-by-Step Solution

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 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.

Step 2: Find the Weight

(a)

 

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

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:

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

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

Step 3: Find the position

(b)

 

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

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

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

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