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When a body is executing pure rolling we know that the point of contact of the body with the ground is at rest with respect to the ground. If that's the case no friction should act as it is stationary.So when a body is rolling down an inclined plane its point of contact is stationary , then how does friction act to cause a torque, as static friction only acts when there is a tendency of relative motion with respect to the ground.

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When a body is executing pure rolling we know that the point of contact of the body with the ground is at rest with respect to the ground.

This is true.

If that's the case no friction should act as it is stationary

This is false. Static friction is a friction force that can act on an object that is not sliding relative to the surface it is touching.

So when a body is rolling down an inclined plane its point of contact is stationary , then how does friction act to cause a torque, as static friction only acts when there is a tendency of relative motion with respect to the ground?

Gravity is attempting to accelerate the body down the incline. The static friction force opposes this. Since the friction is applied at the edge of the body and tangent to it friction has a torque about the center of the body and it starts to roll.

Contrast this with a body rolling on a flat surface. If there are no other horizontal forces then there is nothing for friction to oppose. Therefore, there is no static friction force, hence no torque. The body will continue to roll at a constant speed. However if I then apply a horizontal force, static friction now wants to oppose this. Hence we now have a torque and a change in speed (this problem, discussed here and here, is actually not trivial. You can get different magnitudes and directions of friction depending on the body and the the location and strength of the applied force if you want rolling without slipping).

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False premise. Static friction only acts when there is no motion between the two objects being considered in the system. For instance, a stationary block on an inclined plane is held there by static friction which equals the component of its weight down the plane, $mg\sin{\theta}$. Under such scenarios, the force of static friction is bounded by: $$F_f\leq\mu_sN$$ Where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

You might be thinking of kinetic friction, which can only occur when the two objects in contact have relative motion.

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    $\begingroup$ Okay, i understand. But then if friction acts when the point of contact is stationary then it must also act during rolling on a horizontal plane surface. Then if friction acts it will cause a torque right? So the ball will continue to accelerate by itself? $\endgroup$
    – ag_1902
    Jun 26, 2019 at 3:53
  • $\begingroup$ Let us consider what happens to a rolling object in a plane. Suppose that the plane is elevated by $\theta$ and that the body has an acceleration down the plane of $a$. Then a free-body diagram shows that: $$mg\sin{\theta}-F_f=ma.$$ A horizontal plane is the limiting case $\theta\rightarrow0$. The first term thus vanishes. Furthermore, for a rolling disk: $$a=\frac{2}{3}g\sin{\theta},$$ which can be derived with simple rotational physics. This means the RHS of our first equation also vanishes as $\theta\rightarrow0$. Hence we conclude on a flat plane, the force of friction is $0$. $\endgroup$
    – Andrew Paul
    Jun 26, 2019 at 4:07
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    $\begingroup$ But μN is not zero right? Then by this logic how is friction zero. $\endgroup$
    – ag_1902
    Jun 26, 2019 at 4:19
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    $\begingroup$ Remember that static friction is only bounded by $\mu_sN$. That is, $F_f\leq\mu_sN$. They are not necessarily equal. So it is perfectly possible to have $F_f=0<\mu_sN$. $\endgroup$
    – Andrew Paul
    Jun 26, 2019 at 4:21
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    $\begingroup$ @ag_1902 What mathysics says is true. It is a common introductory physics mistake to assume the static friction force is always equal to $\mu_sN$. The only time equality holds is if you are looking right at the instant where static friction fails and slipping occurs. $\endgroup$
    – BioPhysicist
    Jun 26, 2019 at 4:26

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