A guy-wire supports a pole that is 95 ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground 50 ft from the base of the pole. Determine the horizontal and vertical components (in lb) of the force of tension in the wire if its magnitude is 90 lb. (Round your answers to the nearest integer.)

The horizontal and vertical components (in lb) of the force of tension in the wire is 42 and 80 lb respectively.

Given the following data:

To calculate the horizontal and vertical components (in lb) of the force of tension in the wire:

The horizontal and vertical components of force.

First of all, we would determine the angle of inclination between the pole and the guy-wire by using this formula:

Angle = 62.24°

For the horizontal component of tensional force:

Mathematically, the horizontal component of tensional force is given by this formula:

Tx = 41.92 ≈ 42 lb.

For the vertical component of tensional force:

Mathematically, the vertical component of tensional force is given by this formula:

Tx = 79.64 ≈ 80 lb.

Read more on horizontal component here: link

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2