Equilibrium of Rigid Bodies (Part II)

Introduction

The object is in equilibrium, when sum of all forces operating on this object, is equal to zero. That means that the object has no any forces which make it rotate or being unsteady. According to Newton’s 3^rd law, each action has counteraction which has same force, but directed to opposite direction. Force of reaction, is that kind of force, which keeps an object to be in equilibrium without participation of surrounding forces. As an example to it, the object which lies on the flat surface can serve. The object operates on a surface and a surface operates on object with the same force. Force with which object operates on a surface passes through its centre of gravity and perpendicular to the surface. Force of the reaction also passes through the centre of gravity of the object. Thereby, it is possible to confirm that the object which lies on the flat surface, without surrounding forces, is in equilibrium. Hence, if the force acting on object does not pass through its centre, when object is in equilibrium, the object will start to rotate aside to which force is directed. The reason for it, is the momentum. Momentum can be defined by the formula:

Where is the force that acting to object, and is the distance between centre of gravity of the object and the acting force.

The experiment is made from two different objectives. Idea of the experiment is using the system if objects, which are in equilibrium, is to learn how to use and how balancing is important for. The first objective is to find the mass of the unknown object using balanced system of known objects and one metre ruler. For that part should know at least one known mass, which will be used, to find the mass of the unknown object or objects. The second part of the experiment is to find the mass and the centre of gravity of two irregular shaped bars, using at least two known masses.

Experimental Procedure

Apparatus:

Metre ruler, iron stand, triple beam balance, set of masses, light string, unknown masses and irregular shaped bars.

Object 1

In this object, originally, need to find the mass of the one unknown object using at least one known mass and one metre ruler, which hanged on the iron stand. In my experiment I have used two unknown masses and one known mass. As unknown masses were used pen and block, as the known mas was used mass of the 50 gram. Hence, experiment has been made two times to get enough information for determination of the unknown masses. That means in different times, objects had different places on the ruler. First of all will be shown and explain each part of the Object 1. Then, will be shown Table, which includes all raw data.

Part A (Experiment 1 explanation)

First of all, will be explained how all system were set up. In the beginning iron stand was connected to the triple beam balance. Then, using light string, one metre ruler was connected to the iron stand. After that, one metre was fixed in the point 48.8cm, which is the centre of mass of the ruler. When one metre ruler was in equilibrium, I started to hang objects. One known object and pen were hanged to the right side, and block was hanged to the left side. Then I found for that objects special places, where system was in equilibrium and did not shacked or moved.

The Figure 1 below is showing how experiment was set up.

Figure 1.

 Light String

Block  Pen   Mass

Figure 1 shows only how experiment was set up, but Figure 1 does not show the forces that acting on it. First of all, of solving problems with balanced system, should be used and shown the Free Body Diagram.

Figure 2.

0

The Figure 2 includes how the objects acting on the ruler and shows how far they placed from beginning of the ruler. Where the gravitational force of the block is, is the gravitational force of the pen and is the gravitational force of the known mass. Although, all this forces acting to the ruler. (All positions of the objects were measured from the 0cm).

Part B (Experiment 2 explanation)

Experiment 2 of the Object 1 had same set up as the Experiment 1, but the position of objects is different.

Figure 3, which is below, is a Free Body Diagram of the balanced system, where all objects has different placement.

Figure 3.

Where is the gravitational force of the block, is the gravitational force of the pen and is the gravitational force of the known mass. Although, all this forces acting to the ruler. (All positions of the objects were measured from the 0cm)

Part C (Raw data and calculating)

During the Experiment 1 and Experiment 2, were recorded raw data of the placements of the objects. Raw data was recorded into the Table. Table 1, which is below, shows raw data of the placements of the objects during the experiment.

Table 1.

Distance from 0 cm (cm)	Experiment 1	Experiment 2
COM of ruler	48.8	48.8
Pen ( )	73	85
Block ( )	39	38.2
Known mass ( )	93.2	95.5

The data about the mass of the objects was recorded into the Table 2, which is below.

Table 2.

Mass of the objects (g)	Experiment 1	Experiment 2
Known mass	50	50

Formulas.

To find the mass of unknown mass, will be used the formula of the momentum (1). Hence, our system is in the equilibrium, can be clearly seen that the sum of the forces is equal to zero. Those means, force that acting to the right side, is equal to the force that acting to the left side. In formula form, it will look like:

(Formula 1.1)

In our Experiments that means:

In this formula was used the point of COM of the ruler, because in formula (1) should be used distance between object and COM, but in raw data in the Table 1 said that distance took from 0 cm.

(Therefore, in calculations will not be used the gravitational force of the ruler, because it is fixed in the Centre of Mass).

Calculating for Experiment 1.

For calculation in the Experiment 1, need substitute raw data from Table 1 and Table to the formula (2).

(3)

Now, need to simplify formula (3):

(Formula 4)

Calculating for Experiment 2.

For calculation in the Experiment 1, need to substitute raw data from Table 1 and Table to the formula (2).

(5)

Now, need to simplify formula (5):

(Formula 6)

Calculating the unknown masses using the formulas (4) and (6).

To find the unknown values of masses need to get two simplified formulas from calculating in Experiment 1 and Experiment 2:

Now, for the easiest way of calculation, need to show in terms of

=>

Now, need to minus formula (8) from formula (7). The result of that is:

(Formula 9)

Hence the result which was gotten from (9), could be find the first unknown mass of the pen:

=>

=> =>

=> =>

So, mass of the pan was found. Now substitute the value of the into the formula (7), to find the mass of the second object (block):

=>

The mass of the block was found.

Part D (Calculating Errors)

The original mass of the pen is and mass of the block is .

Compare to experimental result, can be clearly seen that was made errors during the Experiments. Hence, errors should be calculated in the percentage.

Error of mass of the pen:

Error of mass of the block: