A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is µ. Let the mass of the box be m.

(a) At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane?

(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to α θ > ?

(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?

(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a?

0 3 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Long Answer type Questions as classified in NCERT Exemplar

    Explanation -for the box to just starts sliding down mg

    sin θ = t = μ N = μ m g c o s θ

    tan θ = μ

    θ = t a n - 1 ( μ )

    b) when angle of inclination is increased to α > θ

    F1= mgsin α - f = m g s i n α - μ N

    = mg ( s i n α - μ c o s α )

    c) to keep the box stationary, upward force needed F2= m g s i n α +f= mg ( s i n α + μ c o s α )

    d) if the box is to move with an upward acceleration a then upward force needed 

    F3=mg ( s i n α + μ c o s α )+ma

     

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