Answer:
The magnitude of the tension force is [tex]F_T = 66 N[/tex]
Explanation:
The diagram of the question is shown on the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 13.0kg[/tex]
The length of the rod is [tex]L = 1.80 m[/tex]
The angle the cable (string) makes with the rod is [tex]\theta = 80 ^o[/tex]
The acceleration due to gravity is [tex]g = 10 m/s^2[/tex]
Torque is mathematically represented as
[tex]\tau = F d \ sin \theta[/tex]
There are two torques acting on the rod
The first is torque due to gravitational force
This force is mathematically represented as
[tex]F_g = Mg[/tex]
The distance is
[tex]d = \frac{L}{2}[/tex]
The angle is [tex]\theta = 90^o[/tex]
The first torque due to gravitational force
[tex]\tau_1 = Mg * \frac{L}{2} * sin 90[/tex]
[tex]\tau_1 = Mg \ \frac{L}{2}[/tex]
The second is torque due to the tension on the cable
The tension on the string is [tex]F_T[/tex]
Since this force is acting at the other end of the rod , the is from that point to the point of torque(rotation ) is
[tex]d = L[/tex]
The angle which [tex]F_T[/tex] made with the point it is acting on is given as
[tex]\thteta = 80^o[/tex]
The second torque due to tension in the string is
[tex]\tau_2 = F_T Lsin 80[/tex]
At equilibrium
[tex]Mg \ \frac{L}{2} = F_T L sin 80[/tex]
Making [tex]F_T[/tex] the subject of the formula we have
[tex]F_T =\frac{ Mg }{2 sin \ 80 }[/tex]
Substituting values
[tex]F_T =\frac{ 13.0 * 10 }{2 sin \ 80 }[/tex]
[tex]F_T = 66 N[/tex]
