You didn't respond to paisiello2 comment in post 2, but it appears you tried to treat this as a truss at the joint where load P is applied by isolating the joint and assuming all the vert applied load is taken by the cable. But some of it is taken in shear by the lower member. Trust your equilibrium equations when determining support reactions from externally applied loads.Isaac Reed said:
A distributed load in statics refers to a type of force that is spread out over a specific area or length, rather than being concentrated at a single point. It is commonly used to model the weight or pressure of objects that are distributed along a certain surface or structure.
A point load is a single force that is applied at a specific point, while a distributed load is spread out over an area or length. Point loads are usually easier to calculate and model, but distributed loads are more realistic and commonly encountered in real-world situations.
The units of measurement for a distributed load depend on the type of load and the units of measurement used for other quantities in the problem. For example, a distributed load due to weight may be measured in pounds per foot or kilograms per meter.
To calculate the total force of a distributed load, you need to integrate the load function over the length or area where it is applied. This will give you the total force in Newtons or pounds. Alternatively, you can break the distributed load into smaller sections and use the formula for calculating the area or length of a shape to find the total force.
Some common examples of distributed loads include the weight of a person or object on a floor or bridge, the pressure of water or wind on a surface, and the weight of soil or other materials on a structure. Distributed loads are also commonly used in engineering and construction to model the weight of building materials such as steel beams and concrete slabs.