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Question

# A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. The minimum coefficient of static friction required, if the vehicle moves with a speed of 15m/s is given by n8. Then, n is [Take g=10 m/s2]

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Solution

## The turn is banked for a speed of 10 m/s. As speed of the vehicle is 15 m/s i.e. greater than banked speed, therefore vehicle will have a tendency to slip outwards. To prevent it, static friction will act inwards on the banked surface. On applying equilibrium condition along vertical direction, Ncosθ−mg−fsinθ=0 ⇒Ncosθ−fsinθ=mg ...(1) On applying equation of dynamics towards the centre of horizontal circular path: Nsinθ+fcosθ=mv2r ...(2) On mulitplying (1) with −sin θ and (2) with cos θ and adding them gives, −Ncosθsinθ+fsin2θ+Nsinθcosθ+fcos2θ=−mgsinθ+mv2rcosθ ⇒f(sin2θ+cos2θ)=mv2rcos θ−mgsinθ f=mv2rcos θ−mgsinθ ...(3) Since tan θ=v2rg [Condition for banking of road] ⇒tan θ=10220×10 ⇒tan θ=12=t2t From triangle shown in figure, cos θ=2√5 sin θ=1√5 Putting the values in Eq.(3), ⇒f=200×15220×2√5−200×10×1√5 ∵v=15 m/s ⇒f=500√5 Since static friction will act at its limiting value to prevent slipping of vehicle ⇒μN>500√5 ⇒μ×mgcosθ>500√5 ⇒μ>500√5200×10×2√5 ⇒μ>25004000=58 Thus, minimum required coefficient of static friction μ=58 On comparing with μ=n8, we get n=5

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