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M1 Help - Statics of a Particle

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    I'm stuck on this question:

    A particle is in equilibrium on a rough horizontal table. A string is attached to the particle and held at an angle 30 degrees to the horizontal. The tension in the string is T, the frictional force exerted by the table on the particle is F. Find

    a) F given T = 25N

    I'm just totally crap at M1! Its probably really simple to do what I did is draw a diagram and from it got F - mgsin theta = 25 and then I inputted mg as 25 and theta as 30 and ended up wid 37.5 which is completely wrong in the back its 21.7 I don't have a clue how to do it any help will be appreciated and rep will be given.

    Thanks
    S001.
  2. Offline

    Hmm, no. If you draw a diagram, friction is acting horizontally; so for equilibrium you need to resolve the tension component horizontally.
  3. Offline

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    Isn't that what I've done :?
  4. Offline

    No, draw the diagram. I get the impression that your table isn't horizontal, and it should be.
  5. Offline

    No, you've done something quite odd, but the sin(theta) part suggests vertical resolution. For one thing, you don't know if the string's direction is above or below the horizontal. But yeah, don't think vertical, think horizontal. Innit. Perhaps I'll draw a diagram.
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    You'd do me a great favour if you draw the diagram. I have know worked it out by inputting cos instead of sin(theta) but I don't know why this is the case?
  7. Offline

    (Original post by S001)
    You'd do me a great favour if you draw the diagram. I have know worked it out by inputting cos instead of sin(theta) but I don't know why this is the case?
    It's right-angled triangle trigonometry. cos(30) = Adjacent / Hypotenuse; with the adjacent side being the horizontal component of the tension, and the hypotenuse is the tension.

    (For the diagram, I've drawn the string going up. Since we're only interested in the horizontal components, whether it goes up or down is irrelevant.)
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    Thanks for the help
  9. Offline

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    Here are my thoughts on this.....

    The particle is in equilibrium, meaning the forces must balance, vertically and horizontally.

    The frictional force on the table, F can be defined by the equation F=mew*R....mew is the frictional coeffiecient and R is the reaction force (vertically).

    The weight of the particle is mg. The tension in the string is 25N (which will be the same at either end!!). The frictional force, F will act in the opposite direction (horizontally) to T.

    Resolve the Tension in the string. Vertical = 25cos30 degress and Horizontal = 25sin30 degrees.


    Considering Vertical forces - R + 25cos30degress = mg

    Considering Horizontal froces - F=mew*R=25sin30 degrees

    F=25sin30 degress for equilibrium conditions. Therefore, the frictional force is equal to 12.5N.


    ****I may be totally wrong on this but these are only my thoughts*******

    Best of luck,

    Pipsqueek
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    (Original post by pipsqueek)
    Here are my thoughts on this.....

    The particle is in equilibrium, meaning the forces must balance, vertically and horizontally.

    The frictional force on the table, F can be defined by the equation F=mew*R....mew is the frictional coeffiecient and R is the reaction force (vertically).

    The weight of the particle is mg. The tension in the string is 25N (which will be the same at either end!!). The frictional force, F will act in the opposite direction (horizontally) to T.

    Resolve the Tension in the string. Vertical = 25cos30 degress and Horizontal = 25sin30 degrees.


    Considering Vertical forces - R + 25cos30degress = mg

    Considering Horizontal froces - F=mew*R=25sin30 degrees

    F=25sin30 degress for equilibrium conditions. Therefore, the frictional force is equal to 12.5N.


    ****I may be totally wrong on this but these are only my thoughts*******

    Best of luck,

    Pipsqueek
    I think you have the right idea but your cos and sin have been swapped! Think about it, the sin of 30 multiplied by the tension is the vertical component...draw a triangle and check!
  11. Offline

    ReputationRep:
    You are correct, yes it is 25sin30 degrees for the vertical component, silly mistake by me. Thus, the horizontal component is 25cos30 degrees.

    The overall frictional force, F, is equal to 25cos30 degrees = 25*3^2/2=21.7N (1dp)

    Pipsqueek.

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