8.2 Newton's second law of motion

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3: The body is a real three dimensional entity. The force system is concurrent at a common point. But this common point does not coincide with the center of mass. The context of study in this case is also same as that for the case in which force system is not concurrent. We apply Newton’s second law for translation as defined for the center of mass and Newton’s second law for angular motion (we shall define this law at a later stage) for angular or rotational effect.

The discussion so far assumes that the body under consideration is free to translate and rotate. There are, however, real time situations in which rotational effect due to external force is counter-balanced by restoring torque. For example, consider the case of a sliding block on an incline. Application of an external force on the body along a line, not passing through center of mass, may not cause the body to overturn (rotate as it moves). The moment of external force i.e. applied torque may not be sufficient enough to overcome restoring torque due to gravity. As such, if it is stated that body is only translating under the given force system, then we assume that the body is a point mass and we apply Newton’s second law straight way as if the body were a point mass.

Unless otherwise stated or specified, we shall assume that the body is a point mass and forces are concurrent. We shall, therefore, apply Newton’s second law, considering forces to be concurrent, even if they are not. Similarly, we will consider that the body is a point mass, even if it is not. For example, we may consider a block, which is sliding on an incline. Here, “friction force” is along the interface, whereas the “normal force” and “weight” of the block act through center of mass (C). Obviously these forces are not concurrent. We, however, apply Newton’s second law for translation, as if forces were concurrent.

Exercises

A ball of mass 0.1 kg is thrown vertically. In which of the following case(s) the net force on the ball is zero? (ignore air resistance)

(a) Just after the ball leaves the hand

(b) During upward motion

(c) At the highest point

(d) None of above cases

The only force acting on the ball is gravity (gravitational force due to Earth). It remains constant as acceleration due to gravity is constant in the vicinity of Earth. Thus, net force on the ball during its flight remains constant, which is equal to the weight of the ball i.e.

$F=\mathrm{mg}=0.1X10=1\phantom{\rule{2pt}{0ex}}N$

Hence, option (d) is correct.

A pebble of 0.1 kg is subjected to different sets of forces in different conditions on a train which can move on a horizontal linear direction. Determine the case when magnitude of net force on the pebble is greatest. Consider g = 10 $m/{s}^{2}$ .

(a) The pebble is stationary on the floor of the train, which is accelerating at 10 $m/{s}^{2}$ .

(b) The pebble is dropped from the window of train, which is moving with uniform velocity of 10 m/s.

(c) The pebble is dropped from the window of train, which is accelerating at 10 $m/{s}^{2}$ .

(d) Magnitude of force is equal in all the above cases.

In the case (a), the pebble is moving with horizontal acceleration of 10 $m/{s}^{2}$ as seen from the ground reference (inertial frame of reference). The net external force in horizontal direction is :

$⇒{F}_{\text{net}}=\mathrm{ma}=0.01X10=1\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\text{(acting horizontally)}$

There is no vertical acceleration. As such, there is no net external force in the vertical direction. The magnitude of net force on the pebble is, therefore, 1 N in horizontal direction.

In the case (b), the pebble is moving with uniform velocity of 10 m/s in horizontal direction as seen from the ground reference (inertial frame of reference). There is no net force in horizontal direction. When dropped, only force acting on it is due to gravity. The magnitude of net force is, thus, equal to its weight :

$⇒{F}_{\text{net}}=\mathrm{mg}=0.01X10=1\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\text{(acting downward)}$

In the case (c), the pebble is moving with acceleration of 10 $m/{s}^{2}$ as seen from the ground reference (inertial frame of reference). When the pebble is dropped, the pebble is disconnected with the accelerating train. As force has no past or future, there is no net horizontal force on the pebble. Only force acting on pebble is its weight. The magnitude of force on the pebble is :

$⇒{F}_{\text{net}}=\mathrm{mg}=0.01X10=1\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\text{(acting downward)}$

Hence, option (d) correct.

A rocket weighing 10000 kg is blasted upwards with an initial acceleration 10 $m/{s}^{2}$ . The thrust of the blast is :

$\mathrm{\left(a\right)}\phantom{\rule{2pt}{0ex}}1X{10}^{2}\phantom{\rule{2pt}{0ex}}\mathrm{\left(b\right)}\phantom{\rule{2pt}{0ex}}2X{10}^{5}\phantom{\rule{2pt}{0ex}}\mathrm{\left(c\right)}\phantom{\rule{2pt}{0ex}}3X{10}^{4}\phantom{\rule{2pt}{0ex}}\mathrm{\left(d\right)}\phantom{\rule{2pt}{0ex}}4X{10}^{3}$

The net force on the rocket is difference of the thrust and weight of the rocket (thrust being greater). Let thrust be F, then applying Newton’s second law of motion :

${F}_{\text{net}}=F-\mathrm{mg}=\mathrm{ma}$

$⇒F=m\left(g+a\right)=10000X\left(10+10\right)=200000=2X{10}^{5}\phantom{\rule{2pt}{0ex}}N$

Hence, option (b) is correct.

The motion of an object of mass 1 kg along x-axis is given by equation,

$x=0.1t+5{t}^{2}$

where “x” is in meters and “t” is in seconds. The force on the object is :

$\mathrm{\left(a\right)}\phantom{\rule{2pt}{0ex}}0.1\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\mathrm{\left(b\right)}\phantom{\rule{2pt}{0ex}}0.5\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\mathrm{\left(c\right)}\phantom{\rule{2pt}{0ex}}2.5\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\mathrm{\left(d\right)}\phantom{\rule{2pt}{0ex}}10\phantom{\rule{2pt}{0ex}}N$

The given equation of displacement is a quadratic equation in time. This means that the object is moving with a constant acceleration. Comparing given equation with the standard equation of motion along a straight line :

$x=\mathrm{ut}+\frac{1}{2}a{t}^{2}$

We have acceleration of the object as :

$a=5X2=10\phantom{\rule{2pt}{0ex}}m/{t}^{2}$

Applying Newton’s second law of motion :

$F=\mathrm{ma}=1X10=10\phantom{\rule{2pt}{0ex}}N$

Hence, option (d) is correct.

A bullet of mass 0.01 kg enters a wooden plank with a velocity 100 m/s. The bullet is stopped at a distance of 50 cm. The average resistance by the plank to the bullet is :

$\mathrm{\left(a\right)}\phantom{\rule{2pt}{0ex}}10\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\mathrm{\left(b\right)}\phantom{\rule{2pt}{0ex}}100\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\mathrm{\left(c\right)}\phantom{\rule{2pt}{0ex}}1000\phantom{\rule{2pt}{0ex}}N\phantom{\rule{2pt}{0ex}}\mathrm{\left(d\right)}\phantom{\rule{2pt}{0ex}}10000\phantom{\rule{2pt}{0ex}}N$

The plank applies resistance as force. This force decelerates the bullet and is in opposite direction to the motion of bullet. Since we seek to know average acceleration, we shall consider this force as constant force assuming that wooden has uniform constitution. Let the corresponding deceleration be “a”. Then, according to equation of motion,

${v}^{2}={u}^{2}+2\mathrm{as}$

$⇒a=-\frac{{v}^{2}-{u}^{2}}{2s}$

Putting values,

$⇒a=-\frac{{10}^{4}-0}{2X0.5}=-{10}^{4}\phantom{\rule{2pt}{0ex}}m/{t}^{2}$

Applying Newton’s second law of motion :

$⇒F=\mathrm{ma}=0.01X{10}^{4}=100\phantom{\rule{2pt}{0ex}}N$

Hence, option (b) is correct.

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