Irodov Problem 1.61

(a) The force diagram of the problem is depicted in the picture beside. It is easier to solve the problem if we resolve all forces along the directions parallel and perpendicular to the inclined plane. This is shown for each of the two bodies separately in the figure.

Forces acting on :
Let us first consider the direction perpendicular to the inclined plane. There are only two forces acting in this direction - the normal reaction from the surface and the component of gravity that acts on the mass, . There is no acceleration for the mass in this direction and hence we have,




Now, let us consider the forces along the direction of the inclined plane. There are three forces acting in this direction - i) the component of gravity trying to accelerate the mass down , ii) the force of friction opposing the direction of motion and iii) the normal reaction from body 2 to body 1, F pushing downwards along the inclined plane. Suppose that the mass accelerates downwards with an acceleration of w along the inclined plane. We have,






Forces on :
There are two forces acting in the perpendicular direction to the inclined plane - the normal reaction from the inclined plane and the component of force of gravity in a direction perpendicular to the inclined plane . There is no acceleration for body 2 along this direction. Thus, we have,




There are three forces acting in the horizontal direction, i) the component of gravity along the inclined plane , ii) the normal reaction from body 1 to body 2 pushing body 2 up the inclined plane F and, iii) the force of friction that opposes the motion of body 2 given by, . The body also accelerates at a rate w along this direction down the inclined plane. Thus, we have,






Now we can solve for the unknowns F and w. Adding (6) and (3),






Now we can substitute (7) in either (3) or (6) to obtain,





(b) Strictly speaking friction coefficient during motion and during the time when the body is static are usually different values - the friction coefficient in the static position is usually high. In this problem however, both these values are assumed to be the same. The friction force constantly adjusts itself and provide equal and opposite force to oppose motion until the maximum friction it can provide (normal reaction times the coefficient of friction) is reached. Thus, all have to do is set w=0 in (7) and so we have,

Irodov Problem 1.60

The force diagrams for each of the three bodies is depicted in the figure beside.

Forces on :
The tension in the string pulls the mass upwards while the gravitational force pulls it downwards and the net acceleration is w. Thus,



Forces on :
In the vertical direction there are two forces - the normal reaction from the surface and the force of gravity pulling it down. However, there is no downward acceleration for the body. Thus, we have,




Now in the vertical direction there are three forces acting on the mass, the tension from the two strings on either sides of the mass pulling the it in opposite directions (as shown in the figure) and the force of friction that is given by in a direction so as to oppose any motion (the acceleration). Further the mass moves with an acceleration w. Thus, we have,






Forces on :
In the vertical direction there are two forces acting on mass - the normal reaction from the surface acting upwards and the force of gravity acting downwards. There is no acceleration for this mass in the vertical direction. Thus, we have,




In the horizontal direction there are two forces acting on the mass - the tension in the string and the force of friction. The mass accelerates at a rate w. Thus, we have,








Now, we can solve for w as below,












Now we can use (9) and (7) to obtain the value of the tension between the masses as,

Irodov Problem 1.59

First to give some background. An aerostat is a dirigible sort of like a balloon. There's the balloon's buoyant force pushing it up and the force of gravity of the mass attached to the balloon pulling it down. The way a balloon can be made to go up is either by throwing off some mass (ballast mass) attached to the balloon or by heating the air in the balloon.

Let the buoyant force of the aerostat pushing it up be F. Let the mass attached to the balloon be m. Since the balloon is accelerating downwards with at a rate w. We can write the force equations as,




Suppose that we have to throw a mass to make sure that the balloon accelerates upwards at a rate w then once again from newton's laws,

Irodov Problem 1.58

At first the object just turns around the stationary axis AB at and angular velocity oriented along AB. Then, AB itself turns around the axis CD with a constant angular acceleration - the acceleration vector is directed along CD. Now at any instant, the the object is also rotating about CD with an angular velocity vector directed along CD. The net angular velocity vector (obtained by the addition of the two vectors) magnitude is thus given by,.

The object experience two kinds of angular accelerations, i) along CD with magnitude and ii) the other because the angular velocity vector EF itself rotates about CD.

The angle that the vector EF has rotated by time t is given by . Thus, the angular velocity of the particle as a function of time is given by,








This is the last of the problems in the kinematics section of I.E. Irodov. Now we shall move on to the fundamental equation of dynamics.