- #1
Diagram pressure forces over a body resting on an inclined surface
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Homework Help Overview
The discussion revolves around understanding the diagram of pressure forces acting on a body, specifically a prism resting on an inclined surface. Participants are exploring the nature of pressure distribution and its representation in diagrams.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants discuss whether the diagram pressure is always represented as a straight line and consider the implications of forces acting on the center of mass. Questions are raised about scenarios where the pressure might not be a straight line, and requests for additional resources are made.
Discussion Status
The conversation is ongoing, with participants providing insights into the nature of pressure forces and discussing the assumptions involved. There is an acknowledgment of the complexity of the problem, particularly regarding different support scenarios and the need for additional information about the prism's internal structure.
Contextual Notes
Participants note the lack of definitive mathematical proof for the assumptions regarding pressure lines and the variability in reaction forces depending on the support conditions of the body in question.
- #2
rock.freak667
Homework Helper
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...well yes essentially pressure acts in a straight line, normal to a surface. In the diagram, only the component of the force reaction surface (the force normal reaction) would exert a pressure.
- #3
Bob S
- 4,662
- 8
You might want to consider a torque about the CM, because the friction force does not act on CM, while the gravitational force does.
- #4
- #5
tiny-tim
Science Advisor
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orlan2r said:
Hi orlan2r!
You're asking about the line formed by the ends of the vectors representing the reaction force at each point, as shown in the diagrams.
So far as I know, it is assumed that the ends lie along a line, but there is no mathematical way to prove this.
Consider a beam supported at two points … there are two independent equations, from which we can find the two reaction forces.
But if the beam is supported at three points, there are still only two independent equations, but there are three reaction forces.
We can only find those three forces if we include a third equation, which in that case will describe the elasticity of the beam.
Similarly, we usually assume that a table supported on four legs has four equal reaction forces. But that need not be so, as is easily seen by cutting one of the legs … the table can still stand on three legs!
We cannot say what the "diagram pressure" will be for your prism, unless we know the internal structure of the prism.