Question

Derive expression of acceleration of abody On smooth including plane up and down

Answer 1

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Explanation:

The component of mg parallel to surface of incline (downward ) is mg sin(theta). Here, (theta) is angle of incline. Since, body is just to start the motion it experiences limiting static frictional force, f(s) =(mu) mg cos(theta),where (mu) is coefficient of static friction for surface of incline and the body in contact with each other. The equation of motion will be

ma=mg sin(theta)-(mu) mgcos(theta) or

acceleration, a=g[sin(theta)-(mu)cos(theta)].

Thus, when body is just to start motion the acceleration is what we have written. This acceleration remains constant through out the motion.

When sin(theta)=(mu) cos(theta) or (mu)=tan (theta), then, a=0.

This shows that for (theta)= tan^-1(mu), the body will be on the point of just starting to slide . This means that an infinitesimal increase in (theta) will cause acceleration a=g[sin(theta)-(mu)cos(theta)]

HOPE IT HELPS YOU FRIEND...☺

》》Ⱥʀɳɑѵ☃️

Answer 2

$\text {\large \underline { \green{ ☆Answer☆ }}}$

Explanation:

The component of mg parallel to surface of incline (downward ) is mg sin(theta). Here, (theta) is angle of incline. Since, body is just to start the motion it experiences limiting static frictional force, f(s) =(mu) mg cos(theta),where (mu) is coefficient of static friction for surface of incline and the body in contact with each other. The equation of motion will be

ma=mg sin(theta)-(mu) mgcos(theta) or

acceleration, a=g[sin(theta)-(mu)cos(theta)].

Thus, when body is just to start motion the acceleration is what we have written. This acceleration remains constant through out the motion.

When sin(theta)=(mu) cos(theta) or (mu)=tan (theta), then, a=0.

This shows that for (theta)= tan^-1(mu), the body will be on the point of just starting to slide . This means that an infinitesimal increase in (theta) will cause acceleration a=g[sin(theta)-(mu)cos(theta)]

HOPE IT HELPS YOU FRIEND...☺

》》Ⱥʀɳɑѵ☃️