Anonuniform beam 4.50 m long and weighing 1.40 kn makes an angle of 25.0° below the horizontal. it is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it
the center of gravity of the beam is 2.00 m down the beam from the pivot. lighting equipment exerts a 5.00-kn downward force on the lower left end of the beam. find the tension t in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. start by sketching a free-body diagram of the beam.

(Downwards)

(Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

Now force balance in vertical direction

(Downwards)

Force balance in horizontal direction

(Towards Left)

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).

true

explanation: