# 11.2 Newton's second law  (Page 2/3)

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In a similar investigation where the mass is kept constant, but the applied force is varied, you will find that the bigger the force is, the faster the object will move. The acceleration of the trolley is therefore directly proportional to the resultant force. In mathematical terms:

$a\propto F$

Rearranging the above equations, we get $a$ $\propto$ $\frac{F}{m}$ OR $F\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}ma$

Newton formulated his second law as follows:

Newton's Second Law of Motion

If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. The mathematical representation is:

$F=ma.$

## Applying newton's second law

Newton's Second Law can be applied to a variety of situations. We will look at the main types of examples that you need to study.

A 10 kg box is placed on a table. A horizontal force of 32 N is applied to the box. A frictional force of 7 N is present between the surface and the box.

1. Draw a force diagram indicating all the horizontal forces acting on the box.
2. Calculate the acceleration of the box.
1. We only look at the forces acting in a horizontal direction (left-right) and not vertical (up-down) forces. The applied force and the force of friction will be included. The force of gravity, which is a vertical force, will not be included.

2. We have been given:

Applied force F ${}_{1}$ = 32 N

Frictional force F ${}_{f}$ = - 7 N

Mass m = 10 kg

To calculate the acceleration of the box we will be using the equation ${F}_{R}=ma$ . Therefore:

$\begin{array}{ccc}\hfill {F}_{R}& =& ma\hfill \\ \hfill {F}_{1}+{F}_{f}& =& \left(10\right)\left(a\right)\hfill \\ \hfill 32-7& =& 10\phantom{\rule{3.33333pt}{0ex}}a\hfill \\ \hfill 25& =& 10\phantom{\rule{3.33333pt}{0ex}}a\hfill \\ \hfill a& =& 2,5\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\phantom{\rule{3pt}{0ex}}\mathrm{towards the left}\hfill \end{array}$

Two crates, 10 kg and 15 kg respectively, are connected with a thick rope according to the diagram. A force of 500 N is applied. The boxes move with an acceleration of 2 m $·$ s ${}^{-2}$ . One third of the total frictional force is acting on the 10 kg block and two thirds on the 15 kg block. Calculate:

1. the magnitude and direction of the frictional force present.
2. the magnitude of the tension in the rope at T.
1. Always draw a force diagram although the question might not ask for it. The acceleration of the whole system is given, therefore a force diagram of the whole system will be drawn. Because the two crates are seen as a unit, the force diagram will look like this: Force diagram for two crates on a surface
2. To find the frictional force we will apply Newton's Second Law. We are given the mass (10 + 15 kg) and the acceleration (2 m $·$ s ${}^{-2}$ ). Choose the direction of motion to be the positive direction (to the right is positive).

$\begin{array}{c}\hfill {F}_{R}=ma\\ \hfill {F}_{\mathrm{applied}}+{F}_{f}=ma\\ \hfill 500+{F}_{f}=\left(10+15\right)\left(2\right)\\ \hfill {F}_{f}=50-500\\ \hfill {F}_{f}=-450N\end{array}$

The frictional force is 450 N opposite to the direction of motion (to the left).

3. To find the tension in the rope we need to look at one of the two crates on their own. Let's choose the 10 kg crate. Firstly, we need to draw a force diagram: Force diagram of 10 kg crate

The frictional force on the 10 kg block is one third of the total, therefore:

${F}_{f}=\frac{1}{3}×$ 450

${F}_{f}=150\phantom{\rule{4pt}{0ex}}\mathrm{N}$

If we apply Newton's Second Law:

$\begin{array}{ccc}\hfill {F}_{R}& =& ma\hfill \\ \hfill T+{F}_{f}& =& \left(10\right)\left(2\right)\hfill \\ \hfill T+\left(-150\right)& =& 20\hfill \\ \hfill T& =& 170\phantom{\rule{4pt}{0ex}}\mathrm{N}\hfill \end{array}$

Note: If we had used the same principle and applied it to 15 kg crate, our calculations would have been the following:

$\begin{array}{ccc}\hfill {F}_{R}& =& ma\hfill \\ \hfill {F}_{\mathrm{applied}}+T+{F}_{f}& =& \left(15\right)\left(2\right)\hfill \\ \hfill 500+T+\left(-300\right)& =& 30\hfill \\ \hfill T& =& -170\phantom{\rule{4pt}{0ex}}\mathrm{N}\hfill \end{array}$

The negative answer here means that the force is in the direction opposite to the motion, in other words to the left, which is correct. However, the question asks for the magnitude of the force and your answer will be quoted as 170 N.

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