Q61P - University Physics with Modern Physics

Question

Two ropes are connected to a steel cable that supports a hanging weight (Fig. P5.61). (a) Draw a free-body diagram showing all of the forces acting at the knot that connects the two ropes to the steel cable. Based on your diagram, which of the two ropes will have the greater tension? (b) If the maximum tension either rope can sustain without breaking is 5000N , determine the maximum value of the hanging weight that these ropes can safely support. Ignore the weight of the ropes and of the steel cable.

                       

Step-by-Step Solution

Answer

(a)

The first rope will have the greatest tension.

(b) The maximum value of the hanging weight that these ropes can safely support is 6441.83 N.

Step 1: Identification of the given data

The given data is listed below as:

 

• The maximum tension the ropes can contain is, ${\mathrm{T}}_1=5000\mathrm{N}$ .
• The first rope makes an angle of ${\mathrm{\theta }}_1=60°$ with the horizontal.
• The second rope makes an angle of ${\mathrm{\theta }}_2=40°$ with the horizontal.

Step 2: Significance of the tension

Tension is described as the force that the objects exert on each other. The tension force mainly happens with the help of a rope or a string.

Step 3: (a) Determination of the free body diagram

The diagram of the forces acting at the knot has been described below:

 

           

Here in the above diagram, the weight is acting downwards and the tensions of the rope $T_1$ and $T_{\mathit{2}}$ makes and angle of $60°$ and $40°$ with the horizontal. Moreover, the tension T has two components such as $T_{x_1}$ and $T_{{\mathit{x}}_{\mathit{2}}}$ in the x direction and $T_{{\mathit{y}}_{\mathit{1}\mathit{-}\mathit{2}}}$ in the y direction. Furthermore, $T_{\mathit{1}}$ makes an angle of$120°$with the horizontal.

Step 4: (a) Determination of the ropes which will have the greatest tension

As the maximum tension that either ropes can sustain without breaking is $T_{\mathit{1}}\mathit{=}5000\mathrm{N}$, then as the tension is mainly coming from the first rope, hence, it is the tension of the first rope.

 

The summation of all the forces in the direction is zero according to the free body diagram.

 

The equation of the tension of the ropes in the direction is expressed as:

$$\begin{gathered} \sum _{ }{\mathrm{F}}_{\mathrm{x}}={\mathrm{T}}_{{\mathrm{x}}_1}+{\mathrm{T}}_{{\mathrm{x}}_2} \\ =0 \end{gathered}$$ 

            

Here, $\sum _{ }{\mathrm{F}}_{\mathrm{x}}$ is the summation of the forces in the x direction and ${\mathrm{T}}_{{\mathrm{x}}_1}$ and ${\mathrm{T}}_{{\mathrm{x}}_2}$ are the tension in the direction.

 

The above equation can also be expressed as:

 

$$\begin{gathered} {\mathrm{T}}_{{\mathrm{x}}_1}+{\mathrm{T}}_{{\mathrm{x}}_2}=0 \\ {\mathrm{T}}_1\mathrm{cos\theta }+{\mathrm{T}}_2{\mathrm{cos\theta }}_2=0 \end{gathered}$$

 

Here, $T_1$ is the tension in the first rope, is the angle made by the tension in the first rope, $T_2$ is the tension in the second rope and is the angle made by the tension in the second rope.

 

Substitute 5000N for $T_1$ , $120°$ for $\theta$ and $40°$ for ${\theta }_2$ in the above equation.

 

$$\begin{gathered} \left(5000\mathrm{N}\right)\cos 120°+{\mathrm{T}}_2\cos 40°=0 \\ \left(5000\mathrm{N}\right)\cos 120°=-{\mathrm{T}}_2\cos 40° \\ {\mathrm{T}}_2=-3263.51\mathrm{N} \end{gathered}$$

 

Thus, the first rope will have the greatest tension.

Step 5: (b) Determination of the maximum value of the hanging weight

The summation of all the forces in the y direction is zero according to the free body diagram.

 

The equation of the tension of the ropes in the y direction is expressed as:

$$\begin{gathered} \sum _{ }{\mathrm{F}}_{\mathrm{y}}={\mathrm{T}}_{\mathrm{y}1}+{\mathrm{T}}_{\mathrm{y}2} \\ =0 \end{gathered}$$ 

                                                                                                                  …(i)

Here, $T_{y1}$ and $T_{\mathit{y}\mathit{2}}$ are the tension in the y direction.

 

The equation of the weight can be expressed as:

 

$\sum _{ }{\mathrm{F}}_{\mathrm{y}}=w$ 

 

Here, w is the weight hanging from the ropes.

 

 

Substitute w for $\sum _{ }{\mathrm{F}}_{\mathrm{y}}$ in the above equation.

$w=T_{y_1}+T_{y2}$ 

 

The above equation can also be expressed as:

 

               $$\begin{gathered} T_{x_1}+{T_x}_2=w \\ {T}_1\sin \theta +T_2\sin {\theta }_2=w \end{gathered}$$

 

Here, $T_1$ is the tension in the first rope, $\theta$ is the angle made by the tension in the first rope, $T_2$ is the tension in the second rope ${\theta }_2$ and is the angle made by the tension in the second rope.

 

Substitute 5000 N for ${\mathrm{T}}_1$, 3263.51 N for ${\mathrm{T}}_2$ , $120°$ for $\theta$ and $40°$ for ${\theta }_2$ in the above equation.

 

$$\begin{gathered} \mathrm{w}=\left(5000\mathrm{N}\right)\sin 120°+\left(3263.51\mathrm{N}\right)\sin 40° \\ =\left(5000\mathrm{N}\right)\left(0.866\right)+\left(3263.51\mathrm{N}\right)\left(0.642\right) \\ =4330\mathrm{N}+2111.83\mathrm{N} \\ =6427.87\mathrm{N} \end{gathered}$$

 

Thus, the maximum value of the hanging weight that these ropes can safely support is 6427.87 N.