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a crate of mass 194 kg is loaded onto the back of a flatbed truck. the coefficient of static friction between the box and the truck bed is 0.62. what is the smallest radius of curvature that the truck can take without the crate slipping, if the speed with which it is going around a circle is 41 m/s? 276 m



Answer :

The smallest radius of curvature that the truck can take without the crate slipping is 276.38 m.

Given that,

Mass of the crate m = 194 kg

Coefficient of static friction μ = 0.62

Speed of the truck v = 41 m/s

The static force between the box and the truck bed acts as a source for the centripetal force.

So, Centripetal force = Frictional force

(m* v²)/r = μ* m* g

By making r as subject, we have,

r = v²/(μ* g) = 41² / (0.62* 9.81) = 276.38 m

Thus, the smallest radius of curvature that the truck can take without the crate slipping is 276.38 m.

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