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A mass of 10 kg is suspended by a rope o...

A mass of 10 kg is suspended by a rope of length 2.8m from a ceiling. A force of 98 N is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is

A

`30^(@)`

B

`60^(@)`

C

`45^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
`"According to Lami.s theorem"(F_(1))/(sin alpha)=(F_(2))/(sin beta)`
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A mass of 3kg is suspended by a rope of length 2m from the ceiling. A force of 40 N in the horizonal direction is applied at midpoint P of the rope as shown. What is the angle the rope makes with the vertical in equilibrium and the tension in part of string attached to the ceiling? (Neglect the mass of the rope, g=10m//s^(2) )

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Knowledge Check

  • A mass of 10kg is suspended by a rope of length 2.8m from a ceiling. A force of 98N is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is .

    A
    `30^(@)`
    B
    `60^(@)`
    C
    `45^(@)`
    D
    `90^(@)`

  • A mass of 4kg is suspended by a rope of length 4m from a ceiling. A force of 20N in the horizontal direction is applied at the mid-point of the rope as shown in figure. What is the angle which the rope makes with the vertical in equilibrium? Neglect the mass of the rope. Take g=10ms^(-2) .

    A
    `tan^(-1)2`
    B
    `tan^(-1)((1)/(2))`
    C
    `tan^(-1)sqrt(2)`
    D
    `tan^(-1)(1)(sqrt2)`

  • A mass of 6 kg is suspended by a rope of length 2m from a celling. A force of 50N is applied in horizontal direction at the mid point of the rope. What is the angle of the rope with the verticle in equilibrium:-

    A
    `tan^(-1)((4)/(5))`
    B
    `tan^(-1)((5)/(4))`
    C
    `tan^(-1)((5)/(6))`
    D
    None of these

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