Home
Class 12
PHYSICS
When force vec(F)(1),vec(F)(2),vec(F)(3)...

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`

Text Solution

Verified by Experts

The correct Answer is:
B
As initially `vec(F)_(1)+vec(F)_(2)+vec(F)_(3)=0`
If `vec(F)_(1)` is deleted
then `vecF_("net") = vec(F)_(2) +vec(F)_(3)=ma vec(a)=(vec(F)_(2)+vec(F)_(3))/(m)=(vec(F_(1)))/(m)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEM

    RESONANCE|Exercise DPP No.24|9 Videos

  • DAILY PRACTICE PROBLEM

    RESONANCE|Exercise DPP No.25|20 Videos

  • DAILY PRACTICE PROBLEM

    RESONANCE|Exercise DPP No.22|9 Videos

  • CURRENT ELECTRICITY

    RESONANCE|Exercise High Level Problems (HIP)|21 Videos

  • ELECTRO MAGNETIC WAVES

    RESONANCE|Exercise Exercise 3|27 Videos

Similar Questions

Explore conceptually related problems

Three forces (vec(F)_(1), vec(F)_(2), vec(F)_(3)) are acting an a particle moving vertically up with constant speed. Two force vec(F)_(1)=-10hat(j)N , and vec(F)_(1)=-6hat(i)=8hat(j) , N are acting an particle on acting on particle respectively find vec(F)_(3) .

A particle of m=5kg is momentarily at rest at x= at t=. It is acted upon by two forces vec(F)_(1) and vec(F)_(2) . Vec(F)_(1)=70hat(j) N. The direction and manitude of vec(F)_(2) are unknown. The particle experiences a constant acceleration, vec(a) ,in the direction as shown in (figure) Neglect gravity. a.Find the missing force vec(F)_(2) . b. What is the velocity vector of the particle at t=10 s ? c. What third force, vec(F)_(3) is required to make the acceleration of the particle zero? Either give magnitude and direction of vec(F)_(3) or its components.

Knowledge Check

  • When force vec(F_(1)), vec(F_(2)),vec(F_(3))"…..."vec(F_(n)) act on a particle , the particle remains in equilibrium . If vec( F_(1)) is now removed then acceleration of the particle is

    A
    `(vec(F_(1)))/(m)`
    B
    `- (vec(F_(1)))/(m)`
    C
    `- ( vec( F_(2))+vec(F_(3))+"....."vec(F_(n))-vec(F_(1)))/(m)`
    D
    `(vec(F_(1)))/(m)`

  • When forces F_1 , F_2 , F_3 , are acting on a particle of mass m such that F_2 and F_3 are mutually perpendicular, then the particle remains stationary. If the force F_1 is now removed then the acceleration of the particle is

    A
    (a) `F//m`
    B
    (b) `F_2F_3//mF_1`
    C
    (c) `(F_2-F_3)//m`
    D
    (d) `F_2//m`

  • When forces F_(1) , F_(2) , F_(3) are acting on a particle of mass m such that F_(2) and F_(3) are mutually prependicular, then the particle remains stationary. If the force F_(1) is now rejmoved then the acceleration of the particle is

    A
    `F_(1)//m`
    B
    `F_(2)F_(3)//mF_(1)`
    C
    `(F_(2)-F_(3))//m`
    D
    `F_(2)//m`

    • Similar Questions

    Explore conceptually related problems

    Five forces vec(F)_(1) ,vec(F)_(2) , vec(F)_(3) , vec(F)_(4) , and vec(F)_(5) , are acting on a particle of mass 2.0kg so that is moving with 4m//s^(2) in east direction. If vec(F)_(1) force is removed, then the acceleration becomes 7m//s^(2) in north, then the acceleration of the block if only vec(F)_(1) is action will be:

    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.

    Three forces are acting on a particle of mass m initially in equilibrium if the first two forces (vec(R)_(1)and vec(R)_(2)) are perpendicular to each other and suddenly the third force (vec(R)_(3)) is removed, then the magnitude of acceleration of the particle is:

    Two forces vec(F_(1)) and vec(F_(2)) are acting at right angles to each other , find their resultant ?

    A force F_1 acts on a particle so as to accelerate it from rest to a velocity v. The force F_1 is then replaced by F_2 which decelerates it to rest

    Home
    Profile