Q10E - University Physics with Modern Physics

Question

In Fig. E5.10 the weight w is 60.0 N. 

(a) What is the tension in the diagonal string? 

(b) Find the magnitudes of the horizontal forces F₁ and F₂ that must be applied to hold the system in the position shown.

 

Step-by-Step Solution

Answer

(a) The tension in the diagonal string is $\mathbf{85}\mathbf{ }\mathbf{N}$ .

(b) Both horizontal forces to hold the system is $\mathbf{60}\mathbf{ }\mathbf{N}$ .

Step 1: Tension in a string

Given Data:

• The weight of the block, $w=60\mathrm{N}$ .
• The angle of the diagonal string with horizontal, $\beta =45°$ . 

Tension in the string:

The tension in the diagonal string can be found by using Lami’s theorem, and the horizontal forces to hold the block can be found by equating these forces with the horizontal component of tension in the diagonal string.

Step 2: Determine the tension in a diagonal string by using Lami’s theorem (a)

The tension in the diagonal string can be calculated as:

 

$\frac{T}{\sin \alpha }=\frac{w}{\sin \left(180°-\beta \right)}$  

 

Here, $\alpha$ is the angle between the force $F_2$ and weight w, $\beta$ is the angle of diagonal string from horizontal, $w$ is the weight of the block

 

Substitute all the values in the above equation, and we get,

  $$\begin{gathered} \frac{T}{\sin \left(90°\right)}=\frac{60\mathrm{N}}{\sin \left(180°-\left(45°\right)\right)} \\ T=85\mathrm{N} \end{gathered}$$

 

Therefore, the tension in the diagonal string is  $85\mathrm{N}$.

Step 3: Determine the magnitude of horizontal forces by vertical and horizontal equilibrium (b)

The first horizontal force to hold the system is given as:

 

$F_1=T\cos \beta$  

 

Here, $F_1$ is the first force to hold the block, $T$ is the tension in the diagonal string

 

Substitute all the values in the above equation, and we get,

 

$$\begin{gathered} F_1=\left(85\mathrm{N}\right)\times \left(\cos 45°\right) \\ {F}_1=60\mathrm{N} \end{gathered}$$  

 

The second horizontal force to hold the system is given as:

 

$F_2=T\sin \beta$  

 

Here $F_2$ is the second force to hold the block, $\beta$ is the angle for the second force

 

Substitute all the values in the above equation, and we get,

 $$\begin{gathered} F_1=\left(85\mathrm{N}\right)\left(\sin 45°\right) \\ {F}_1=60\mathrm{N} \end{gathered}$$

 

Therefore, both horizontal forces to hold the system is $60\mathrm{N}$.