A flexible drive belt runs over a frictionless pulley as shown in figure. The pulley is rotating freely about the vertical axis passing through the centre ‘O’ of the pulley. The vertical axis is fixed on the horizontal smooth surface. The mass per unit length of the drive belt is $$1$$ $$kg/m$$ and tension in the drive belt is $$8N$$. The speed of the drive belt is $$2m/s$$. Find the net normal force applied by the belt on the pulley in newton.

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Infinity Learn · Accepted Answer

Consider an element of mass $$dm=$$λRdθ at angle θ from $$reference.2Tsin$$⁡dθ$$2$$−$$dN=$$λRdθ$$v2R($$λ is linear mass density of belt $$)$$∴Tdθ−$$dN=$$λ$$v2d$$θDue to symmetry , dNsinθ component will cancel out and dNcosθ adds up. ∴ So total normal $$=$$ ∫$$-$$π$$2$$$$π$$$$2dNcos$$⁡θ$$=$$∫$$-$$π$$2$$$$π$$$$2Tcos$$⁡θdθ−λ$$v2$$∫$$-$$π$$2$$$$π$$$$2cos$$⁡θdθ⇒$$N=Tsin$$$$π$$$$2--$$$$sin$$$$π$$$$2-$$λ$$v2sin$$$$π$$$$2--$$$$sin$$$$π$$$$2=2T-2$$λ$$v2$$⇒$$N=2(8)-2(1)22=8N$$

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