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Pressure and force analysis

  1. Sep 28, 2011 #1
    I don't understand what makes us analyise a certain force in to two perpendicular components
    see this page
    why do we calculate pressure in this image from the relation P = F cos q/A
    why not P=F/A?? Is this beacuse pressure is always perpendicular?and if so why is pressure perpendicular,when the force makes an angle with the normal?
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 28, 2011 #2
    If you apply a force at some angle to a surface as shown in your picture, that force has two components.

    One component is perpendicular to the area. This is called the normal stress or direct stress or pressure. This is Fcos([itex]\theta[/itex]) /A

    The other component is parallel to the area and is called the shear stress. It is not a pressure, since it does not 'press' on the area, but drags along the surface. This is Fsin([itex]\theta[/itex]/A

    F is the resultant of these two forces the normal force and the shear force.

    go well
  4. Sep 29, 2011 #3
    What is a shear stress?
    I don't understand why is F the resultant of those two forces
  5. Sep 29, 2011 #4
    Good morning, misr.

    When you resolve a force in some direction, as you have done in your diagram, you get two components (in 2 dimensions).

    I assume you understand this since you showed Fcos([itex]\theta[/itex])

    The component you showed is called the normal force. This acts perpendicular to the area.

    The other one is called the shear force. This acts parallel to the area.

    Do you understand this so far?
  6. Sep 29, 2011 #5
    Yeah,I understand this
    but is this normal force due to gravity or something else?
    because the object is pressing upon the surface,so the surface reacts by this normal force?correct?
  7. Sep 29, 2011 #6
    I really have no idea what the source of F is, you drew the diagram after all.

    A few basic ideas.

    In mechanics we need to distinguish between Force and stress.

    Stress is Force divided by area.

    I think we both know that Force has a specific line of action and a point of application.

    We distinguish between body forces such as gravity, which act throughout a body and surface forces which are externally applied by some agent (eg the tension in a string).
    We consider the body force of gravity to conform to my rule above by saying that this force is acts at the centre of gravity and is called the weight. This force cannot directly exert a pressure it is an internal force.

    If a block is resting on a (horizontal) table its weight exerts a downward pressure on the table. This pressure equals the weight divided by the area of the block sitting on the table.
    I have sketched this in fig1.

    As noted above this is also called the stress imposed on the table by the block.

    Since the weight acts perpendicular to the table that is all there is to it.

    Now suppose I replaced the block with a light plate and a spring compressed so as to exert a force on the plate equal to W, the weight of the block. Although the source of the force on the table is quite different the effect is the same ie the pressure is the same, so long as the spring force is exerted vertically. I have sketched this in Fig2.

    However we tilt the spring to apply its force at an angle, as shown in Fig3.
    What do you think the pressure on the table would be if the angle was 90 degrees as shown in fig4?
    Obviously there is no vertical force now being applied to the table via the plate. There is, however a still force being applied.
    This is called the shear force (in this case it is also the friction force).

    If we return to Fig3 and apply our spring force at some intermediate angle we have two components, one vertical and one horizontal. This is a more general situation.

    does this help?

    Attached Files:

    • pre1.jpg
      File size:
      12.3 KB
  8. Sep 30, 2011 #7
    I don't understand this.Is the plate fixed to the table and you are pulling the spring?

    why is the pressure the same for both fig(1) and (2)

    why do we define the pressure as the vertical force per unit area?why not the net force per unit area?
  9. Sep 30, 2011 #8
    Now why do you suppose I said the spring was compressed?

    I do hope you are reading this fully.

    No the plate just sits on the table.
    I said it is a light plate so that means it has no weight of its own.

    The spring pushed down against the plate.

    Well I did say the force the spring pushed down with was set to be the same as the weight of the block.

    Now try reading my post again and see if it makes more sense.
  10. Oct 5, 2011 #9
    Certainly,It makes more sense now
    I guess I understand what are you trying to say,the pressure becomes less when it is not perpendicular

    Now I want to ask another question:
    Why do we define pressure as "AVERAGE" force acting normally on unit area at this point
    as in the page provided above?
  11. Oct 5, 2011 #10
    Yes that's right.
    If two equal forces are acting (on equal surfaces) one perpendicular and one at an angle - the perpendicular force exerts more pressure than the angled one.

    Most people just accept what they are told, but you are obviously a thinking person so here is some extra detail.

    The terms 'normal stress' and 'pressure' refer to the same physical phenomenon.
    'Normal stress' is usually used in conection with solids and 'pressure' in connection with fluids.
    We sometimes talk about pressure in connection with solids when we are considering contact stresses between two solids for instance 'foundation pressure' or 'bearing pressure'.

    In the first sketch I have a 1kg weight sitting on a block of concrete, which is much bigger than the weight.
    On the surface (section AA) where the weight is sitting the weight is concentrated only over the area of the contact surface, not over the whole area of AA.
    As we go deeper into the concrete the 1kg spreads out over a wider and wider part of the concrete until we can say that the weight exterts an average pressure of 1kg divided by the area of the concrete block at section CC.
    At intermediate section BB the pressure exerted by the weight is intermediate between that at AA and CC.

    So what would happen if the block of concrete extended much further?

    Well in sketch 2 I have shown the foundation pressure under a building of weight W. You can see a series of 'bowls of soil' that get larger and larger in area as we get further from the building. So W is distributed over an increasing area and the pressure gradually diminishes over these increasing areas.

    Back to fluids, for although the pressure is the same in all directions at a point in a fluid, it can still vary from point to point.
    So in sketch 3 I have shown the steadily increasing pressure of the water on the back of a dam. This increases linearly from nothing at the surface to a maximum at the base. As a result I have shown a triangle of forces.
    I do not know if you have yet covered centre of gravity?
    The 'average' pressure is the pressure at the centre of gravity (properly called the centre of pressure) of the triangle. The force on the dam equals this average pressure times the wetted area of the back of the dam.

    go well

    Attached Files:

    Last edited: Oct 5, 2011
  12. Oct 6, 2011 #11
    I can't imagine this :(
    In fig (1)As we go deeper ,the pressure on a certain surace is not affected according to the relation P=h*rho*g
    where h is the height of the block
  13. Oct 6, 2011 #12
    A concrete block is not a fluid.

    However the concrete at each section also experiences the pressure due to the weight of the concrete above it, just like a fluid.

    But I am only considering the effect of the 1kg weight - or if you like the extra effect of the 1kg weight.
  14. Oct 6, 2011 #13
    This one is okay,but if so how could we calculate the total pressure on the back of the dam,if each point on the back of the dam has a different pressure?
  15. Oct 6, 2011 #14
    You are right..sorry for this stupidness
    but I still can't imagine
  16. Oct 6, 2011 #15
    There is no such animal as 'total pressure' - There is only average pressure over a whole area or specific pressure at a point. Pressure does not 'add up' over the area.

    You are thinking of Force.

    If I exert 10 pascal over 1 square metre and 10 pascal over another half a square metre there is no total pressure, just a pressure of 10 pascal acting.

    The force, however is 10 newton in the first case and 15 in the second

    Have a good look at the response by Halls of Ivy to your question about the fluid in the tank, it is similar to my dam example.

    It is fundamental and very important to distinguish between force and pressure.
  17. Oct 6, 2011 #16
    It's coming up to 1 am where I am and, as you can see from my passport photo, I desperately need my beauty sleep so we will have to continue this another day, but please come back and confirm you have conquered the difference (and link as they are intimately connected) between force and pressure.

    go well
  18. Oct 9, 2011 #17
    Well,the main problem here is that i can't imagine how pressure is distributed over a larger area on going deeper..
    Isn't the contact area the same?

    What are you talking about?What is "the centre of gravity"?

  19. Oct 9, 2011 #18
    Is this a language problem or a physics problem? We really need to know the extent of your knowledge to be able to help as a centre of gravity is a pretty basic concept in physics.

    I also said more, as did Redbelly in another thread, that it is fundamental you understand the difference between force and pressure.

    This is fine if we need to start there (or even further back), but progress cannot be made without it.
  20. Oct 11, 2011 #19
    It's a physics problem
    and I think I know the difference between force and pressure,pressure is the force divided by area,anything else?
  21. Oct 11, 2011 #20
    The 'Centre of Gravity' of any body is the point through which all the weight of that may be be considered to act as a single concentrated force.

    For instance consider a canon ball. We take the weight to be one force - say W, acting vertically downwards at the geometric centre of the ball.

    Things may get much more complicated however - take for instance an L shaped metal bracket.

    Now the weight of a body is just the sum of all the weights of the individual small elements added up

    W = sum of welements

    Are you familiar with sigma notation?

    W = Ʃw

    W and w are, of course, forces.

    We can add forces from a different cause that are distributed over a body, area or volume in the same way to obtain a 'centre of force' for that particular force.

    The 'Centre of Pressure' is just such a calculation.
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