Question

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

Answer 1

As mentioned in the question, the truck moves in a circular path which means centripetal force acts on the system. Friction force also acts on the system. Then the value given can be substituted in the formula .

Formula used:
Centripetal force $=\dfrac{m{{v}^{2}}}{r}$
Coefficient of friction $\mu =\dfrac{F}{N}$

Complete step-by-step answer:
According to the question truck carries a box having mass $m$ & moves with velocity $v$. The truck is moving on a flat horizontal surface, so frictional force will act on the system.
As frictional force & centripetal force are acting towards the centre of the curvature, the maximum velocity frictional force also acts towards the center of the curvature and it should be maximum that provides necessary sufficient centripetal force.
Given-

& m=50kg \\\ & \mu =0.3 \\\ & r=0.7 \\\ \end{aligned}$$ Coefficient of friction can be given as - $$\mu =\dfrac{F}{N}$$ Where $${{F}_{1}}$$ is the frictional force acting on a body and $$N$$ is the normal force. Centripetal force is given by- $${{f}_{c}}=\dfrac{m{{v}^{2}}}{r}$$ Therefore, $${{F}_{1}}={{F}_{c}}$$ $$\mu N=\dfrac{m{{v}^{2}}}{r}$$ Here, normal force $$N=mg$$ Rearranging the terms, $${{v}^{2}}=\dfrac{\mu Nr}{m}$$ Substituting the values $$\begin{aligned} & {{v}^{2}}=\dfrac{0.3\times 50\times 10}{50\times 27} \\\ & v=9m/s \\\ \end{aligned}$$ Therefore, with speed 9 m/s the box will not slide from the truck even if following a circular path. **Additional Information:** The force which is used to keep an object in a curved path is known as centripetal force. Centripetal force has the tendency to accelerate the object by changing its direction. **Note:** As truck moves in circular path centripetal force will act. For calculating maximum velocity frictional force equalizes the centripetal force and both these forces act towards the center of curvature.