Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.
[tex]\Sigma \tau = 0[/tex]
We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]
Doing the summation using their respective lever arms:
[tex]0 = L Tsin\theta - dF_g[/tex]
[tex]dF_g = LTsin\theta[/tex]
Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:
[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]
Now, let's solve for 'T'.
[tex]T = \frac{dMg}{Lsin\theta}[/tex]
Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]
