Given,
The mass of the box, m=24 kg
The acceleration due to gravity, g=10 m/s²
The angle made by slanted rope, θ=37°
As the system is in equilibrium, there is no net force acting on the mass.
(a) Therefore, the y-component of the tension T₁ will be equal to the weight of the box.
That is,
[tex]T_1\sin \theta=mg[/tex]
Thus,
[tex]T_1=\frac{mg}{\sin \theta}[/tex]
On substituting the known values,
[tex]\begin{gathered} T_1=\frac{24\times10}{\sin 37^{\circ}} \\ =398.8\text{ N} \end{gathered}[/tex]
Thus the tension T₁ is 398.8 N
(b) As the system is in equilibrium the net force is zero in x-direction too.
Thus the x-component of tension T₁ will be equal to T₂
That is,
[tex]T_2=T_1\cos \theta[/tex]
On substituting the known values,
[tex]\begin{gathered} T_2=398.8\times\cos 37^{\circ} \\ =318.5\text{ N} \end{gathered}[/tex]
Thus the tension T₂ is 318.5 N
