A truck is carry a box of mass $m = 50 k g$ on its flat horizontal rough surface with coefficient of friction $μ = 0.3$ . It is crossing a circular track of radius $27 m$ . What is the maximum speed of the truck so that the box does not slide from the truck while moving on the circular path?

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Solution

Here, for maximum velocity friction force towards center of curvature should be maximum which should provide sufficient centripetal force
$f_{l} = F_{c}$ $\Rightarrow μ N = m v^{2} / r$
$0.3 \times 50 \times 10 = 50 \times v^{2} / r$ (here $N = m g$ )
$\Rightarrow v^{2} = 3 \times 27$
this gives , $v = 9 m / s$

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A truck is carry a box of mass m=50 kg on its flat horizontal rough surface with coefficient of friction μ=0.3. It is crossing a circular track of radius 27 m. What is the maximum speed of the truck so that the box does not slide from the truck while moving on the circular path?

Here, for maximum velocity friction force towards center of curvature should be maximum which should provide sufficient centripetal forcefl=Fc ⇒μN=mv2/r0.3×50×10=50×v2/r (hereN=mg)⇒v2=3×27this gives , v=9m/s

A truck is carry a box of mass $m = 50 k g$ on its flat horizontal rough surface with coefficient of friction $μ = 0.3$ . It is crossing a circular track of radius $27 m$ . What is the maximum speed of the truck so that the box does not slide from the truck while moving on the circular path?