8.12 Unbalanced force system (application)  (Page 23/2)

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to unbalanced force system. The questions are categorized in terms of the characterizing features of the subject matter :

• Nature of force
• Motion under gravity
• Combined motion
• Constrained motion

Nature of force

Problem 1 : A pendulum bob oscillates between extreme positions in a vertical plane. The string connected to the bob is cut, when it is at one of the extreme points. What would be the trajectory of the bob after getting disconnected?

Solution : This question emphasizes two important aspects of force. First, force has no past or future. It means that when force is removed, its effect is over. Second, if the body is at rest, then motion (velocity) of the body is in the direction of applied force.

When the string is cut, pendulum bob is at the extreme position and its velocity is zero. The tension in the string immediately disappears as the string is cut. At that instant, the only force on the pendulum bob is force due to gravity i.e. "mg", which acts in vertically downward direction. Thus, pendulum bob has straight trajectory in the downward direction.

Motion under gravity

Problem 2 : With what minimum acceleration a person should slide down a hanging rope, whose breaking strength is ${\mathrm{2/3}}^{\mathrm{rd}}$ of his weight?

Solution : The breaking strength of rope is less than the weight of the person. As such, the rope will break, if the person simply hangs on the rope. It is, therefore, required that the person should climb down with acceleration greater than a minimum value so that tension in the rope is less than its breaking strength.

Let the person is sliding down with an acceleration, "a". The forces on the person are (i) tension in upward direction and (ii) weight of the person in downward direction. The free body diagram of the person sliding down the rope is shown here.

$$\begin{array}{l}\sum F_y=mg-T=ma\\ \Rightarrow T=mg-ma\end{array}$$

According to question, the limiting value of tension in the rope is :

$$\begin{array}{l}T=\frac{2mg}{3}\end{array}$$

Putting the value of limiting tension in the equation of analysis, we have :

$$\begin{array}{l}\Rightarrow \frac{2mg}{3}=mg-ma\\ \Rightarrow a=g-\frac{2g}{3}=\frac{g}{3}\end{array}$$

Problem 3 : Two small spheres of same dimensions “A” and “B” of mass “M” and “m” (M>m) respectively are in free fall from the same height against air resistance “F”. Then, which of the spheres has greater acceleration?

Solution : The free body diagrams of the two bodies are shown in the figure. Let $a_1$ and $a_2$ be the accelerations of “A” and “B” blocks respectively. The force analysis in y-direction is as given here :