Respuesta :
Explanation:
It is given that,
Mass of the box, m = 100 kg
Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.
From the attached figure, the x and y component of forces is given by :
[tex]T_{1x} =-T_1 cos (20)[/tex]
[tex]T_{2x} = T_2 cos (40)[/tex]
[tex]mg_x = 0[/tex]
[tex]T_{1y} = T_1 sin (20)[/tex]
[tex]T_{2y} = T_2 sin (40)[/tex]
[tex]mg_y= -mg[/tex]
Let [tex]R_x[/tex] and [tex]R_y[/tex] is the resultant in x and y direction.
[tex]R_x=-T_1 cos (20)+T_2 cos (40)+0[/tex]
[tex]R_y=T_1 sin(20)+T_2 sin(40)-mg[/tex]
As the system is balanced the net force acting on it is 0. So,
[tex]-T_1 cos (20)+T_2 cos (40)+0=0[/tex].............(1)
[tex]T_1 sin(20)+T_2 sin(40)-100\times 9.8=0[/tex]..................(2)
On solving equation (1) and (2) we get:
[tex]T_1=866.86\ N[/tex] (tension on the left rope)
[tex]T_2=1063.36\ N[/tex] (tension on the right rope)
So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.
