Answer:
B.4638 N
Explanation:
We are given that
Distance of cable from its pivot end=1.25 m
[tex]\theta=25^{\circ}[/tex]
Length of beam=5 m
Half length of beam=[tex]\frac{5}{2}=2.5 m[/tex]
Mass =m=100 kg
We have to find the magnitude of tension in the cable .
According to question
[tex]Tsin25\times 1.25=100\times 9.8\times 2.5[/tex]
[tex]T=\frac{100\times 9.8\times 2.5}{sin25\times 1.25}[/tex]
[tex]T=4638 N[/tex]
Hence,option B is true.