Answer
Verified
400.8k+ views
Hint: We can find the value of coefficient of friction by finding the ratio of the resistive force and the value of normal or perpendicular force. The force of friction always acts in the direction opposite to the direction of intended or actual motion.
Formula used:
$T=Fr\sin \theta $
Where $T$is the torque, $F$ is the force, $r$ is the radius and $\theta $ is the angle between $F$ and lever arm
Complete step by step answer:
Coefficient of friction is described as the ratio of the force of friction between two bodies and the force pressing them together. It is a value that represents the relationship between two objects and the normal force or reaction between the objects. Coefficient of friction is of two types, Static and Kinetic coefficient of friction. Static coefficient of friction is applied to the stationary bodies while Kinetic coefficient is applied to the bodies in motion. Stationary or motionless objects experience more friction than those which are already in motion. We can say that it requires more force to put an object in motion than to sustain them in motion.
In the above question we are given that the rod does not slip on the floor, meaning we can apply the equations to equate all the forces as the rod is in equilibrium.
The forces acting on the rod are – gravitational force, applied force and the torque about the point of intersection of gravitational and applied force.
$\begin{align}
& \text{Gravitational force = }mg \\
& \text{Applied force = }F \\
& \text{Torque = }T \\
& \text{Angle of torque = }\alpha \\
\end{align}$
Balancing all the forces and torque acting on the rod,
$\sum\limits_{{}}^{{}}{T=Nl\cos \alpha -fl\left( \dfrac{1}{\sin \alpha }+\sin \alpha \right)}=0$
Where $N$i s the normal force and $f$ is the force of friction
The moment arm for normal force $N$ is $l\cos \alpha $ while for friction it is $l\left( \dfrac{1}{\sin \alpha }+\sin \alpha \right)$
From above equation we get,
\[\begin{align}
& f=N\left( \dfrac{\cos \alpha \times \sin \alpha }{1+{{\sin }^{2}}\alpha } \right)=N\left( \dfrac{\cos \alpha \times \sin \alpha }{2{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha } \right) \\
& f=N\left( \dfrac{\cos \alpha \times \sin \alpha }{2\tan \alpha +\cot \alpha } \right) \\
\end{align}\]
Also, the force of friction cannot exceed $\mu N$, we have
$\mu \ge \dfrac{1}{2\tan \alpha +\cot \alpha }$
For minimum value of$\mu $, $2\tan \alpha +\cot \alpha $should be maximum
$\begin{align}
& \dfrac{d(2\tan \alpha +\cot \alpha )}{dx}=0 \\
& 2{{\sec }^{2}}\alpha -\cos e{{c}^{2}}\alpha =0 \\
& \tan \alpha =\dfrac{1}{\sqrt{2}} \\
\end{align}$
We get$\mu =\dfrac{1}{2\sqrt{2}}$
Comparing with$\dfrac{1}{x\sqrt{x}}$,
Value of $x=2$
Note:
While calculating the torque, always choose a point where maximum lines of forces are passing so that the number of resultant torques will be minimum. For no confusion, choose clockwise direction as positive and anti-clockwise as negative.
Formula used:
$T=Fr\sin \theta $
Where $T$is the torque, $F$ is the force, $r$ is the radius and $\theta $ is the angle between $F$ and lever arm
Complete step by step answer:
Coefficient of friction is described as the ratio of the force of friction between two bodies and the force pressing them together. It is a value that represents the relationship between two objects and the normal force or reaction between the objects. Coefficient of friction is of two types, Static and Kinetic coefficient of friction. Static coefficient of friction is applied to the stationary bodies while Kinetic coefficient is applied to the bodies in motion. Stationary or motionless objects experience more friction than those which are already in motion. We can say that it requires more force to put an object in motion than to sustain them in motion.
In the above question we are given that the rod does not slip on the floor, meaning we can apply the equations to equate all the forces as the rod is in equilibrium.
The forces acting on the rod are – gravitational force, applied force and the torque about the point of intersection of gravitational and applied force.
$\begin{align}
& \text{Gravitational force = }mg \\
& \text{Applied force = }F \\
& \text{Torque = }T \\
& \text{Angle of torque = }\alpha \\
\end{align}$
Balancing all the forces and torque acting on the rod,
$\sum\limits_{{}}^{{}}{T=Nl\cos \alpha -fl\left( \dfrac{1}{\sin \alpha }+\sin \alpha \right)}=0$
Where $N$i s the normal force and $f$ is the force of friction
The moment arm for normal force $N$ is $l\cos \alpha $ while for friction it is $l\left( \dfrac{1}{\sin \alpha }+\sin \alpha \right)$
From above equation we get,
\[\begin{align}
& f=N\left( \dfrac{\cos \alpha \times \sin \alpha }{1+{{\sin }^{2}}\alpha } \right)=N\left( \dfrac{\cos \alpha \times \sin \alpha }{2{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha } \right) \\
& f=N\left( \dfrac{\cos \alpha \times \sin \alpha }{2\tan \alpha +\cot \alpha } \right) \\
\end{align}\]
Also, the force of friction cannot exceed $\mu N$, we have
$\mu \ge \dfrac{1}{2\tan \alpha +\cot \alpha }$
For minimum value of$\mu $, $2\tan \alpha +\cot \alpha $should be maximum
$\begin{align}
& \dfrac{d(2\tan \alpha +\cot \alpha )}{dx}=0 \\
& 2{{\sec }^{2}}\alpha -\cos e{{c}^{2}}\alpha =0 \\
& \tan \alpha =\dfrac{1}{\sqrt{2}} \\
\end{align}$
We get$\mu =\dfrac{1}{2\sqrt{2}}$
Comparing with$\dfrac{1}{x\sqrt{x}}$,
Value of $x=2$
Note:
While calculating the torque, always choose a point where maximum lines of forces are passing so that the number of resultant torques will be minimum. For no confusion, choose clockwise direction as positive and anti-clockwise as negative.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE