91影视

Given Data:

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

Tension in the string:

The tension in the diagonal string can be found by using Lami鈥檚 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.

The tension in the diagonal string can be calculated as:

$\frac{T}{\sin 伪}=\frac{w}{\sin \left(180掳-尾\right)}$

Here, $伪$is the angle between the force $F_2$and weight w, $尾$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,

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

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

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

$F_1=T\cos 尾$

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,

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

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

$F_2=T\sin 尾$

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

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

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

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