Question

(II) Find the tension in the two cords shown in Fig. 9–52. Neglect the mass of the cords, and assume that the angle is 33°, and the mass m is 190 kg.

Short answer

The tension in the horizontal cord is 2867 N, and the tension in the inclined cord is 3419 N.

Step-by-step solution

Concepts

In equilibrium, the net force in the x and y directions should be zero.For this problem, find the component of the tension of the inclined cord, compare the force components in both horizontal and vertical directions.

Given data

The mass is $m=190\mathrm{k}\mathrm{g}$.

The angle is $\theta ={33}^{\circ }$.

Let $T_1$ and $T_2$ be the tensions in the cords.

Calculation

The free-body diagram of the problem is given below.

The condition of the equilibrium for the horizontal forces is

$T_2=T_1\cos \theta$ . … (i)

The condition of the equilibrium for the vertical forces is

$\begin{array}{c}{T}_1\sin \theta =mg\\ T_1\sin {33}^{\circ }=\left(190\mathrm{k}\mathrm{g}\right)\times \left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\\ T_1=3419\mathrm{N}\end{array}$.

Now, substituting the value of $T_1$ in equation (i),

$\begin{array}{c}{T}_2=\left(3419\mathrm{N}\right)\cos {33}^{\circ }\\ =2867\mathrm{N}\end{array}$.

Hence, the tension in the horizontal cord is 2867 N, and the tension in the inclined cord is 3419 N.