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A mass of 4 kg rest on a horizontal plan...

A mass of 4 kg rest on a horizontal plane . The plane is gradually inclined until an angle `0 = 15^(@)` with the horizontal and the mass just begins to slide . What is the coefficient of static friction between the block and the surface ?

Text Solution

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Here, `m = 4 kg , theta = 15^(@) , mu_(s) = ?`
As the mass just begins to slide at this angle of inclination , therefore , `theta = 15^(@) = alpha`,
The angle of repose . ` mu_(s) = tan alpha = tan 15^(@) = 0 . 27 `
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Knowledge Check

  • A block of mass 4kg rests on a rough horizontal plane. The plane is gradually inclined until at an angle theta=60^(@) , with the vertical, the mass just begins to side. What is the coefficient of static friction between the block and the plane?

    A
    `sqrt(3)`
    B
    `1/2`
    C
    `(1)/sqrt(2)`
    D
    `(1)/sqrt(3)`

  • A block of mass 20 kg is placed on a rough horizontal surface. When a force of 80 N is applied at an angle of 30°with the horizontal, the block just begins to slide. What is the coefficient of static friction ? (g= 10m//s^2 )

    A
    0.253
    B
    0.433
    C
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    D
    0.75

  • A block of mass 5 kg is placed on a rough inclined plane. The inclination of the plane is gradually increased till the block just begins to slide down. The inclination of the plane is than 3 in 5. The coefficient of friction between the block and the plane is (Take, g = 10m//s^(2) )

    A
    `(3)/(5)`
    B
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    C
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    D
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