Irodov Problem 1.80

If there were no friction, as the system is accelerated towards the left, the mass on the inclined plane would climb up the inclined plane. However, the presence of friction binds them together and moves the mass towards the left along with the inclined plane. After a point, when the friction is not sufficient to hold the mass anymore, it will start to climb up the plane.
Let the mass be m and let the normal reaction offered from the inclined plane to the mass be N.


Forces on the mass m: Let us resolve the forces acting on the mass in directions parallel and perpendicular to the surface of the inclined plane. Along the direction normal to the plane, there are two forces acting, i) the component of gravity and ii) the normal reaction from the surface N. Since the entire system accelerates towards the left with an acceleration w, the component of acceleration along the normal direction is given by as shown in the figure. Thus we have,




In the direction parallel to the inclined plane there are two forces acting on the mass, i) the force of friction opposing the slipping of the mass upwards relative to the inclined plane Nk and ii) the component of gravity pulling it down the inclined plane. The component of acceleration of the mass, as shown in the figure is given by . Thus we have,



Now we can solve for w from (1) and (2) to get,

Irodov Problem 1.79

Let the mass of each of the two bodies be m and let the tension in the string be T. Further let the minimum acceleration required to keep mass from moving be w. Now let us consider the forces acting on each of the masses.

Forces on body 1: There are two forces acting on this body in the vertical direction, i) the normal reaction from the body A and ii) the force of gravity as shown in the figure. There is no acceleration of this mass along the vertical direction. Hence we have,



In the horizontal direction, there are two forces acting on the mass, i) the tension in the string T and ii) the force of friction given by . The mass accelerates at a rate w. Hence, we have,



From (2) and (1) we have,

T - mgk = mw (3)

Forces on body 2: There is only one force acting on the body in the horizontal direction - the normal reaction from the surface which is responsible for pushing it forward at a rate w. Thus, we have,



There are three forces acting on the body in the vertical direction, i) the force of friction opposing the motion of body 2 , ii) the force of gravity pulling it down mg and iii) tension T pulling it up. The mass does not accelerate along this direction. Hence , we have,



From (4) and (5) we have,

mg - T - mwk = 0 (6)
From (3) and (6) we have,

mg - mgk - mw - mwk = 0 or,
g(1-k) = w(1+k) or,