(1) α < g/μ
(2) α > mg/μ
(3) α > g/μm
(4) α ≥ g/μ
When the cart moves forwards with acceleration α, the block of mass m presses against the vertical surface of the cart with a horizontal force of magnitude mα. Therefore, the normal force acting on the block is mα and the maximum frictional force trying to prevent the block from falling is μmα. This must be greater than or equal to the weight mg of the block.
Therefore we have
μmα ≥ mg from which α ≥ g/μ
(2) A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if
(1) r ≥ μg/ω2
(2) r = μgω2
(3) r < ω2/μg
(4) r ≤ μg/ω2
Let the mass of the coin be m. The coin will revolve with the record if the frictional force which supplies the centripetal force (required for the circular motion of the coin) is sufficient. Since the maximum value of the frictional force is μmg, we have
μmg ≥ mrω2
Therefore r ≤ μg/ω2.
You will find more questions (with solution) on friction on this site. One of the posts on friction is here
Similar useful posts in this section can be seen here
No comments:
Post a Comment