91Ó°ÊÓ

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.

The mass is \(m = 190\;{\rm{kg}}\).

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

Let \({T_1}\) and \({T_2}\) be the tensions in the cords.

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

\(\begin{array}{c}{T_1}\sin \theta = mg\\{T_1}\sin {33^ \circ } = \left( {190\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\{T_1} = 3419\;{\rm{N}}\end{array}\).

Now, substituting the value of \({T_1}\) in equation (i),

\(\begin{array}{c}{T_2} = \left( {3419\;{\rm{N}}} \right)\cos {33^ \circ }\\ = 2867\;{\rm{N}}\end{array}\).

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