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    Question is, a particle of m kg slides down a smooth plane inclined at an angle of 30 degrees to horizontal. Find the acceleration down the plane.

    Seems like no informaton is given.

    I have got to
    F = MA
    so mgsin30 = MgA but thats it?
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    (Original post by SunDun111)
    Question is, a particle of m kg slides down a smooth plane inclined at an angle of 30 degrees to horizontal. Find the acceleration down the plane.

    Seems like no informaton is given.

    I have got to
    F = MA
    so mgsin30 = MgA but thats it?
    Yh your on the right lines. I guess the question is a lot simpler than it seems.

    You have F=ma
    mgsin30=ma >>>>>>>>> its just m because its mass
    sin30=0.5
    so 0.5mg/m=a
    so a=0.5g=4.9ms-2
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    Well, resolving component of weight acting down the slope (the only force acting) is  mg \sin 30^{\circ} .
     F=ma \Rightarrow mg\sin 30^{\circ}=ma . I think you can do the rest?
    I didn't read properly that you have already wrote this down, the mass on either side of the equation cancel as they are exactly the same.
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    (Original post by B_9710)
    Well, resolving component of weight acting down the slope (the only force acting) is  mg \sin 30^{\circ} .
     F=ma \Rightarrow mg\sin 30^{\circ}=ma . I think you can do the rest?
    I didn't read properly that you have already wrote this down, the mass on either side of the equation cancel as they are exactly the same.
    if the mass cancel isnt it sin 30 = a so a is 0.5? but thats not the case
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    (Original post by SunDun111)
    if the mass cancel isnt it sin 30 = a so a is 0.5? but thats not the case
    F=ma, not mga
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    Yeah got it guys thanks
 
 
 
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