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The calculation shown is not about the centroid of the bar is located; it's about where the centroid of the triangular load is located from the left support.goldfish9776 said:Homework Statement
why 2/3 is required in this calculation as circled ? I have no idea.
I know that 4.5/2 is the distance of the centroid of the bar from point A .
Homework Equations
The Attempt at a Solution
well , can you explain what the 2/3 means here?SteamKing said:The calculation shown is not about the centroid of the bar is located; it's about where the centroid of the triangular load is located from the left support.
goldfish9776 said:well , can you explain what the 2/3 means here?
2 x 4.5 is the force ... then why 1/2 appear in the calculations?SteamKing said:The centroid of a right triangle is located 2/3 of the length of the base from the pointy end.
https://en.wikipedia.org/wiki/List_of_centroids
Is 2 kN/m x 4.5 m total the force? Wouldn't that be true only if the 2 kN/m distributed load was applied evenly over the entire length of the beam?goldfish9776 said:2 x 4.5 is the force ... then why 1/2 appear in the calculations?
you mean the different points at entire length of rod have different forces? if so, why shouldn't the force = 2x4.5x2/3 ?SteamKing said:Is 2 kN/m x 4.5 m total the force? Wouldn't that be true only if the 2 kN/m distributed load was applied evenly over the entire length of the beam?
goldfish9776 said:you mean the different points at entire length of rod have different forces? if so, why shouldn't the force = 2x4.5x2/3 ?
since the force involved only from point A to centroid ...
ok , why The Total Load = (1/2)*w*L is located at (2/3) of the length of the beam from the end where w = 0.SteamKing said:You've apparently missed some basic instruction in how distributed loads are represented. Here are two commonly encountered types of distributed loadings:
Unlike with a concentrated load, the individual arrows in the figures on the left do not represent individual loads; they are merely diagrammatic. The equivalent concentrated load and its location are given in the corresponding figures to the right.
The first type is the uniformly distributed load. This is a loading which has a constant amount of force applied per unit length along the beam. In the diagram, this load is w kN/m, for example.
The Total Load = w * L and its center is located at L/2 from one end of the beam.
The second type is the uniformly varying load. This is a loading which has a zero amount of force applied per unit length at the right end of the beam, and this loading increases in a linearly increasing fashion until it reaches say w kN/m at the left end of the beam.
The Total Load = (1/2)*w*L and its center is located at (2/3) of the length of the beam from the end where w = 0.
goldfish9776 said:ok , why The Total Load = (1/2)*w*L is located at (2/3) of the length of the beam from the end where w = 0.
why not w*L as in the first case ?
so , at the distance 2L/3 from the position where w= 0 , the force only become wL/2 ?SteamKing said:Because most of the total load is located toward the end opposite of the end of the beam where w = 0.
If you took each beam as loaded in the diagrams, and placed a fulcrum under the location of the center of the total load, the beam would be balanced, which indicates that the moments of the load on either side of the fulcrum produce equal and opposite rotation of the beam. The moments cancel out, the rotations cancel out, and the beam is balanced around the point.
It's like playing on a see-saw: as long as the load on each end of the seesaw is the same and is located the same distance away from the middle, the see-saw is balanced. Add some more load to one side, and the see-saw wants to sink on that side.
I can't believe you've never tried to balance anything, a ruler, a stick, a bat, a baton.
goldfish9776 said:so , at the distance 2L/3 from the position where w= 0 , the force only become wL/2 ?
Is there any mathematical proof for this ? it's hard for me to fully understand it
The equilibrium of rigid bodies refers to the state in which a body is at rest or moving with a constant velocity, with no net force or torque acting on it. This means that the body is balanced and all the forces acting on it are in equilibrium.
Static equilibrium occurs when a body is at rest or moving at a constant velocity, while dynamic equilibrium occurs when a body is moving with a constant acceleration. In static equilibrium, the forces acting on the body are balanced, while in dynamic equilibrium, the forces acting on the body may be unbalanced, but the resulting motion is constant.
In order for a body to be in equilibrium, two conditions must be met: the sum of all the forces acting on the body must be equal to zero, and the sum of all the torques acting on the body must also be equal to zero.
The concept of equilibrium of rigid bodies is used in various fields, such as engineering, architecture, and physics. It is used to analyze the stability of structures, design structures that can withstand external forces, and understand the behavior of objects in motion.
Examples of equilibrium of rigid bodies in everyday life include a book resting on a table, a ladder leaning against a wall, and a person standing still. In all of these situations, the forces acting on the objects are balanced, resulting in a state of equilibrium.