Chapter 9, Problem 12P

Find the tension in the two cords shown in Fig. 9-52. Neglect the mass of the cords, and assume that the angleθ is 33⁰and the mass m is 190 kg.

To determine

The tension in two cords.

Answer to Problem 12P

Solution:

The tension in vertical cord vertical is $3418.77$ and horizontal is $2220.18\text{N}$ .

Explanation of Solution

Given:

Mass of block is 190 kg

Angle, $\theta =33°$

Formula Used:

From second law of motion, equilibrium condition is:

  ${\displaystyle \sum F=0}$

Calculation:

The vertical component of the tension in cord is equal to weight of the box.

Weight $=190\text{kg}\times \text{9}\text{.8=1862N}$

  $\begin{array}{l}{F}_y=T_1\times \sin 33°\\ T_1\times \sin 33°=1862\text{N}\\ T_1=1862/\sin 33°=3418.77\end{array}$

To prevent the box from moving left or right, the horizontal component of the tension in cord must be equal to the tension in cord.

Horizontal force

  $\begin{array}{c}{F}_x=1862/\cos {33}^{°}\\ =2220.18\text{N}\end{array}$

This is tension in horizontal cord

Conclusion:

Hence, the tension in vertical cord vertical is $3418.77\text{N}$ and horizontal is $2220.18\text{N}$.