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Three forces bar(F(1)), bar(F(2)) and ba...

Three forces `bar(F_(1)), bar(F_(2)) and bar(F_(3))` are simultaneously acting on a particle of mass 'm' and keep it in equilibrium. If `bar(F_(1))` force is reversed in direction only, the acceleration of the particle will be.

A

`bar(F_(1))//m`

B

`2bar(F_(1))//m`

C

`-bar(F_(1))//m`

D

`-2bar(F_(1))//m`

Text Solution

Verified by Experts

The correct Answer is:
D
Under equilibrium condition `vec(F_(1))+vec(F_(2))+vec(F_(3))=0`
`vec(F_(1))=-(F_(1)+F_(2)), a=(-F_(1)+F_(2)+F_(3))/(m)`
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Knowledge Check

  • Three forces vecF_(1),vecF_(2) and vecF_(3) are simultaneously acting on a particle of mass 'm' and keep it in equlibrium. If vecF_(1) force is reversed in direction only, the acceleration of the particle will be.

    A
    `barF_(1)//m`
    B
    `2barF_(1)//m`
    C
    `-barF_(1)//m`
    D
    `-2barF_(1)//m`

  • When force vec(F)_(1),vec(F)_(2),vec(F)_(3) are acting on a particle of mass m , the particle remains in equilibrium. If the force vec(F)_(1) is now removed then the acceleration of the particle is :

    A
    `vec(F)_(1)//m`
    B
    `vec(F)_(1)//m`
    C
    `vec(F)_(2)-vec(F)_(3)//m`
    D
    `vec(F)_(2)//m`

  • Which of the following sets of forces acting simultaneously on a particle keep it in equilibrium?

    A
    `3N,5N,10N`
    B
    `4N,5N,12N`
    C
    `2N,6N,5N`
    D
    `5N,8N,1N`

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