Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.
![\Sigma \tau = 0](https://tex.z-dn.net/?f=%5CSigma%20%5Ctau%20%3D%200)
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:
![\tau = r \times F](https://tex.z-dn.net/?f=%5Ctau%20%3D%20r%20%5Ctimes%20F)
Doing the summation using their respective lever arms:
![0 = L Tsin\theta - dF_g](https://tex.z-dn.net/?f=0%20%3D%20L%20Tsin%5Ctheta%20%20-%20dF_g)
![dF_g = LTsin\theta](https://tex.z-dn.net/?f=dF_g%20%3D%20LTsin%5Ctheta)
Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:
![tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7BH%7D%7BL%7D%5C%5C%5C%5Ctan%5E%7B-1%7D%28%5Cfrac%7BH%7D%7BL%7D%29%20%3D%20%5Ctheta%5C%5C%5C%5Ctan%5E%7B-1%7D%28%5Cfrac%7B1.70%7D%7B2.200%7D%29%20%3D%2037.69%5Eo)
Now, let's solve for 'T'.
![T = \frac{dMg}{Lsin\theta}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7BdMg%7D%7BLsin%5Ctheta%7D)
Plugging in the values:
![T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B%280.700%29%284.00%29%289.8%29%7D%7B%282.200%29sin%2837.69%29%7D%20%3D%20%5Cboxed%7B20.399%20N%7D)
Answer:2 amperes
Explanation:
Voltage=120v
Total resistance=15+15+30
Total resistance=60 ohms
Current=voltage ➗ resistance
Current=120 ➗ 60
Current=2 amperes
Answer-
Cono
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