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Question: A uniform rod of length L rests against a smooth roller as shown in Fig. Find the friction coefficie...

A uniform rod of length L rests against a smooth roller as shown in Fig. Find the friction coefficient between the ground and the lower end if the minimum and that the rod makes with the horizontal is θ.

A

μ = lcos⁡θsin⁡2θ2h−lcos⁡2θsin⁡θ\frac{l\cos\theta\sin^{2}\theta}{2h - l\cos^{2}\theta\sin\theta}

B

μ = lsin⁡θcos⁡2θ2h−lcos⁡θsin⁡2θ\frac{l\sin\theta\cos^{2}\theta}{2h - l\cos\theta\sin^{2}\theta}

C

μ = lsin⁡θcos⁡θ2h−lcos⁡θsin⁡θ\frac{l\sin\theta\cos\theta}{2h - l\cos\theta\sin\theta}

D

None

Answer

μ = lcos⁡θsin⁡2θ2h−lcos⁡2θsin⁡θ\frac{l\cos\theta\sin^{2}\theta}{2h - l\cos^{2}\theta\sin\theta}

Explanation

Solution

Balance horizontal force μN2 = N1 sin θ

Balance vertical forces N1 cos θ + N2 = mg

or N1(cos⁡θ+sin⁡θμ)\left( \cos\theta + \frac{\sin\theta}{\mu} \right) = mg

or N1 =μmgμcos⁡θ+sin⁡θ=mg\frac{\mu mg}{\mu\cos\theta + \sin\theta} = mg

Apply ∑τ = 0 about point A.

N1 sin θh + N1 cos θh cot θ = mg 1/2 cos θ

hN1 [sin2θ + cos2θ]= mglcos⁡θsin⁡θ2\frac{mgl\cos\theta\sin\theta}{2}

hμmgμcos⁡θ+sin⁡θ=mgl2\frac{h\mu mg}{\mu\cos\theta + \sin\theta} = \frac{mgl}{2} cos θ sinθ

2μh = μlcos2θ sin θ + l cos θ sin2θ

μ(2h – lcos2θ sinθ) = l cos θ sin2θ

or μ = lcos⁡θsin⁡2θ2h−lcos⁡2θsin⁡θ\frac{l\cos\theta\sin^{2}\theta}{2h - l\cos^{2}\theta\sin\theta}