Question

Regarding the direction of frictional force, which of the following statements is correct?
A. For \[0 \le \theta \le {90^ \circ }\], it always acts leftwards
B. For \[0 \le \theta \le {60^ \circ }\] it acts leftwards and for \[{60^ \circ } \le \theta \le {90^ \circ }\] it acts rightwards
C. It becomes zero at a certain angle \[\theta\] between \[{0^ \circ }\] to \[{90^ \circ }\]
D. None of the above

Answer 1

Hint: When solving questions like these, keep in mind the relationships between angle of friction and angle of repose, and also know how to incorporate those particular concepts into questions. The value of angle of friction depends on the material and nature of surfaces in contact.

Complete step by step solution:

Writing equation of motion (in \[x\] direction)
\[F\cos \theta - {F_r} = ma\] ……(equation 1)
Frictional force is the force that resists motion when a surface of one object comes in contact with the surface of another. Due to friction the output ratio to input ratio is decreased or the mechanical advantage of the machine is sometimes reduced, for which there are provisions made to lessen the friction. There are many types of frictional forces, like static friction, kinetic friction to name a few types.
Torque can be explained as the rotational equivalent of linear force. It is also referred to as moment, moment of force, rotational force, or turning effect. Just as a linear force is a push or pull, a torque can be thought of as a twist to an object around a specific axis. Another way to understand torque can be as a product of magnitude of force and the perpendicular distance of the line of action of a force from the axis of rotation.
Frictional torque is the torque caused by the frictional force that occurs when two objects who are in contact move. Like all other torques, it is a rotational force and can be measured in Newton-metre \[N - m\].
Torque equation for this question can be made as;
\[\Rightarrow \]\[2{F_r}.R(clockwise) + F\cos \theta .R(anticlockwise) + F\sin \theta .R(anticlockwise) = 2m{R^2}\alpha\]
\[\Rightarrow \]\[2{F_r} - F\cos \theta - F\sin \theta = 2m{R^2}\alpha\]
\[\Rightarrow \]\[2{F_r} - F\cos \theta - F\sin \theta = 2ma\] since \[(a = R\alpha )\] ……(2)
Using (equation 1) in (equation 2)
$ \Rightarrow 2(F\cos \theta -ma)-F\cos \theta -F\sin \theta =2ma $
$ \Rightarrow F\cos \theta -F\sin \theta =4ma $
$ \Rightarrow F(\cos \theta -\sin \theta )=4ma $
$ \Rightarrow a=\dfrac{F}{4m(\cos \theta -\sin \theta )} $
$ \Rightarrow F\cos \theta -{{F}_{r}}=\dfrac{mF}{4m(\cos \theta -\sin \theta )} $ Or
$ \Rightarrow F\cos \theta -\dfrac{F}{4(\cos \theta -\sin \theta )}={{F}_{r}} $
$ \Rightarrow {{F}_{r}}=\dfrac{F(4{{\cos }^{2}}\theta -4\sin \theta \cos \theta -1)}{4(\cos \theta -\sin \theta )} $
\[\Rightarrow \]\[{F_r} = \dfrac{F}{4}\dfrac{{(4{{\cos }^2}\theta - 2\sin 2\theta - 1)}}{{(\cos \theta - \sin \theta )}}\]
At \[\theta = {0^ \circ }\] \[{F_r}\] will be maximum.
\[\therefore {{F}_{r}}(\max )=\dfrac{3F}{4}\]
So to conclude we can say that for any angle in between \[{0^ \circ }\] and \[{90^ \circ }\] the direction of frictional force acting against it will be directed leftwards always.
Therefore, the correct answer will be option (A).

Note: Like friction caused by torque, we have other types of friction like fluid friction which is the frictional force that obstructs the flow of fluid. Examples of such friction can be to drop a ball in a bucket full of water, and as a result see water splashing out of the bucket.