JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Laws of Motion

A block of mass

m

is on an inclined plane of angle

θ

. The coefficient of friction between the block and the plane is

μ

and

tan θ > μ

. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive.

As P is varied from

P_{1} = mg (sin θ - μ cos θ)

P_{2} = mg (sin θ + μ cos θ)

, the frictional force

f

versus P graph will look like

(A)(see structure)
✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)

Answer: (A)

Solution

P = P_{1} = m g (sin θ - μ cos θ)

, the block is on the verge of sliding down, so friction

f = μ m g cos θ

acts up the plane. As

P

increases, friction decreases linearly (

f = m g sin θ - P

) until

P = m g sin θ

, where

f = 0

. For

P > m g sin θ

, friction reverses direction (acts down the plane) and increases as

f = P - m g sin θ

until

P = P_{2}

, where

f = μ m g cos θ

again. The graph shows a V-shape with vertex at

(m g sin θ, 0)

, with slopes

- 1

and

+ 1

. Option (D) matches this behavior.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Laws of Motion

Solution

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