Answer:
The tension in the rope is 281.60 N.
Explanation:
Given that,
Length = 3.0 m
Weight = 600 N
Distance = 1.0 m
Angle = 60°
Consider half of the ladder,
let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.
[tex]Y+F=600[/tex]....(I)
[tex]X=T[/tex].....(II)
On taking moment about base
[tex]X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0[/tex]
Put the value into the formula
[tex]X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0[/tex]
[tex]3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30[/tex]
[tex]1.598T=450-1.5(600-F)[/tex]....(III)
We need to calculate the force for ladder
[tex]2F=600\trimes 2[/tex]
[tex]F=600\ N[/tex]
We need to calculate the tension in the rope
From equation (3)
[tex]1.598T=450-1.5(600-600)[/tex]
[tex]1.598T=450[/tex]
[tex]T=\dfrac{450}{1.598}[/tex]
[tex]T=281.60\ N[/tex]
Hence, The tension in the rope is 281.60 N.