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Question

A sphere is rotating between two rough inclined walls as shown in figure. Coefficient of friciton between each wall and the sphere is 13. If f1 amd f2 are the friction forces at P and Q, then f1f2 is

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Solution

The correct option is **A** 4√3+1

Let μ be the friction coefficient between the sphere and each wall. FBD of the sphere is given below. f1 and f2 will act in the directions opposite to tendency of slipping.

Net force on the sphere in horizontal direction is zero.

∴N1cos60∘+μN2cos60∘=N2cos30∘+μN1cos30∘

⇒ N1+μN2=√3(N2+μN1)

⇒ N1(1−√3μ)=N2(√3−μ)

∴ N1N2=√3−μ1−√3μ

Substituting μ=13, we get

N1N2=√3−131−√33=3√3−13−√3=1+4√3

Hence, f1f2=μN1μN2=1+4√3

Let μ be the friction coefficient between the sphere and each wall. FBD of the sphere is given below. f1 and f2 will act in the directions opposite to tendency of slipping.

Net force on the sphere in horizontal direction is zero.

∴N1cos60∘+μN2cos60∘=N2cos30∘+μN1cos30∘

⇒ N1+μN2=√3(N2+μN1)

⇒ N1(1−√3μ)=N2(√3−μ)

∴ N1N2=√3−μ1−√3μ

Substituting μ=13, we get

N1N2=√3−131−√33=3√3−13−√3=1+4√3

Hence, f1f2=μN1μN2=1+4√3

Why this question? This question will help students to grasp the concept of wedge constraint and friction on a rotating wheel. |

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