Irodov Problem 1.68

The force diagram is shown beside. There are three forces acting in the vertical direction, i) the force of gravity mg, ii) the normal reaction from the surface N and iii) the component of the force F acting along the vertical direction . Before the body lifts off the floor, there is no acceleration of the body in the vertical direction, thus,



At the instant when the body looses contact with the ground, the normal reaction N must become 0 (since there is no contact with the ground). Thus, setting N=0 in (1) we for lift-off condition we have,






Now let us consider the motion of the body along the horizontal direction. There is only one force acting on the body along this direction - the component of F namely . If the instantenous acceleration of the body is w, then,
















Again from (3), if x is the distance traveled by the mass at any given instant, we can write,

Irodov Problem 1.67

The figure beside depicts the force diagram. Let us try and find the tension T in the string in the problem. Let the mass be m, and further suppose that the mass is being accelerated along the upwards slope at an rate w. Further let us resolve the forces along the directions parallel and perpendicular to the direction of the inclined plane.

Along the perpendicular direction, there are three forces - i) the normal reaction from the surface of the inclined plane N, ii) the component of gravity and iii) the component of the tension in the string . There is no component of acceleration for the mass along the perpendicular direction. Thus, we have,



Along the direction parallel to the incline, we have 3 forces acting on the mass - i) the component of gravity trying to pull the mass down the inclined plane, ii) the component of tension pulling the mass upwards and iii) the force of friction that opposes the motion of the mass acting downwards (to oppose the upward motion) Nk. We have assumed that the mass accelerates at a rate w and so we have,








Thus, the minimum in tension T can be determined by maximizing . We can find the extrema (maxima or minima) of by differentiating it with respect to and equating this to 0. Thus we have,







Now for this value of to be the maxima its second derivative must be negative. The second derivative is given by,




which is negative throughout 0 to 90 degrees. Thus, the above obtained value of corresponds to the minimum tension in the string.

The actual value of this minimum tension in the string can determined by using this obtained value of in (3) as,







Further, the value of tension will be minimized at w=0 (the mass is pulled up at constant speed) , thus we obtain,

Irodov Problem 1.66

The force diagram of the body on the inclined plane is depicted in the figure beside. We shall resolve the forces acting on the mass along parallel and perpendicular directions to the slope of the inclined plane. Let the mass of the body be m.

Along the direction perpendicular to the mass there are two forces acting - i) the normal reaction from the surface N and ii) the component of gravity acting along the perpendicular direction . There is no component of acceleration along this direction and so we have,



Along the direction parallel to the inclined plane there are two forces acting, i) the component of gravity along the parallel direction and ii) the force of friction that opposes the motion of the mass Nk. If w is the acceleration of the mass then, the dynamics of the mass is given by,








The total length traveled by the mass can be determined by elementary trigonometry as .
Let the total time taken by the mass to reach the base be t. Then,








Thus, t is minimized when is maximized. The condition for extrema (maxima or minima) is given by,







To ascertain whether indeed this is the maxima we must check that the second derivative is negative. The second derivative is given by,










Now from (5) it is clear that and in this interval is positive. Thus, the second derivative of is negative. Thus (5) corresponds to the minima.