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MCQMediumJEE 2023Friction (Static, Kinetic, Rolling)

JEE Physics 2023 Question with Solution

Consider a block kept on an inclined plane (inclined at 45∘45^\circ) as shown in the figure. If the force required to just push it up the incline is 22 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane (μ\mu) is equal to

A small block is shown resting on a rough inclined plane making an angle of 45 degrees with the horizontal.
  • A

    0.330.33

  • B

    0.600.60

  • C

    0.250.25

  • D

    0.500.50

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A block is on a rough incline of angle 45∘45^\circ. Let F1F_1 be the force required to just push it up the incline and F2F_2 be the force required to just prevent it from sliding down.

Find: The coefficient of friction μ\mu.

Free body diagram for pushing the block up the 45 degree incline, showing forces F1, friction f, normal N, mg sin45 and mg cos45.

For impending upward motion, friction acts down the plane.

F1=mgsin⁡45∘+f=mgsin⁡45∘+μNF_1 = mg \sin 45^\circ + f = mg \sin 45^\circ + \mu N

Using N=mgcos⁡45∘N = mg \cos 45^\circ,

F1=mgsin⁡45∘+μmgcos⁡45∘F_1 = mg \sin 45^\circ + \mu mg \cos 45^\circ

Since sin⁡45∘=cos⁡45∘=12\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}},

F1=mg2(1+μ)F_1 = \frac{mg}{\sqrt{2}}(1 + \mu)
Free body diagram for preventing downward sliding on the 45 degree incline, showing forces F2, friction upward, normal N, mg sin45 and mg cos45.

For just preventing downward sliding, friction acts up the plane.

F2=mgsin⁡45∘−f=mgsin⁡45∘−μNF_2 = mg \sin 45^\circ - f = mg \sin 45^\circ - \mu N

So,

F2=mgsin⁡45∘−μmgcos⁡45∘F_2 = mg \sin 45^\circ - \mu mg \cos 45^\circ

Hence,

F2=mg2(1−μ)F_2 = \frac{mg}{\sqrt{2}}(1 - \mu)

Given that F1=2F2F_1 = 2F_2,

mg2(1+μ)=2mg2(1−μ)\frac{mg}{\sqrt{2}}(1 + \mu) = 2\frac{mg}{\sqrt{2}}(1 - \mu) 1+μ=2−2μ1 + \mu = 2 - 2\mu 3μ=13\mu = 1 μ=13=0.33\mu = \frac{1}{3} = 0.33

Therefore, the value of the coefficient of friction is 0.330.33. The solution states the correct option is B, although the listed value 0.330.33 appears in option A. Using the solution as authority, the answer is marked as B with this discrepancy noted.

Using symmetry at $$45^\circ$$

Given: The incline angle is 45∘45^\circ, so the components mgsin⁡45∘mg \sin 45^\circ and mgcos⁡45∘mg \cos 45^\circ are equal.

Find: μ\mu.

At 45∘45^\circ,

mgsin⁡45∘=mgcos⁡45∘=mg2mg \sin 45^\circ = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}

So the two limiting forces become

F1=mg2(1+μ),F2=mg2(1−μ)F_1 = \frac{mg}{\sqrt{2}}(1 + \mu), \qquad F_2 = \frac{mg}{\sqrt{2}}(1 - \mu)

Using F1=2F2F_1 = 2F_2,

1+μ=2(1−μ)1 + \mu = 2(1 - \mu) 1+μ=2−2μ1 + \mu = 2 - 2\mu μ=13=0.33\mu = \frac{1}{3} = 0.33

Therefore, the coefficient of friction is 0.330.33.

Common mistakes

  • Taking friction in the same direction in both cases is incorrect. When the block is about to move up, friction acts down the plane; when it is about to slide down, friction acts up the plane. Reverse the friction direction according to impending motion.

  • Using N=mgN = mg on an incline is wrong because the normal reaction equals only the perpendicular component of weight. Here, use N=mgcos⁡45∘N = mg \cos 45^\circ.

  • Forgetting that sin⁡45∘=cos⁡45∘=12\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} can spoil the simplification. Evaluate both components correctly before applying the given ratio F1=2F2F_1 = 2F_2.

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