# Investigate how the weight of an object affects the force required to overcome friction.

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Introduction

PHYSICS COURSEWORK: THE EFFECT MASS HAS IN RELATION TO STATIC AND DYNAMIC FRICTION

AIM:

To investigate how the weight of an object affects the force

required to overcome friction.

PLANNING:

## Background information

Friction, in mechanics, is the resistance to the sliding, rolling, or flowing motion of a body in relation to another body with which it is in contact. In any solid the molecules show internal friction. This form of friction is the force that causes any oscillating object, such as a piano string or a tuning fork, to stop oscillating. Internal friction in liquids and gases is called viscosity.

There is also external friction and there are two kinds of this, sliding friction and rolling friction. In sliding friction, the resistance is caused by the interference of irregularities on the two surfaces. In rolling friction the resistance is caused by the interference of small deformations formed as one surface rolls over another (see figures 1, 2 and 3).

In both forms of friction molecular attraction between the two surfaces causes some resistance. It is said that the pressures at the contact points of two surfaces are very high, and it is thought that the molecules are pushed into such close proximity that the attractive forces between them weld the surfaces together at these points (see figure 4).

The friction between two objects is at a maximum just before they begin to move in relation to each other and less when the objects are in motion. The maximum value of friction is called static friction and when one of the objects move across the surface of the other the value of friction is called dynamic friction. Static and dynamic friction are said to be directly proportional to the weight acting down on a particular surface.

Figure 1: Sliding Friction

Middle

0.6 / 0.6 / 0.5 0.57

2.548

1.25 / 1 / 1 1.08

1 / 1 / 1 1

3.528

1.6 / 1.6 / 1.5 1.57

1.25 / 1.25 / 1.25 1.25

4.508

1.7 / 2 / 2 1.9

1.5 / 1.5 / 1.6 1.53

5.488

2.5 / 2.4 / 2.5 2.47

2 / 1.8 / 2 1.93

6.468

3.1 / 3 / 3 3.03

2.25 / 2.1 / 2.5 2.28

7.448

3.5 / 3.6 / 3.4 3.5

2.75 / 2.5 / 2.6 2.62

8.428

4.1 / 4.3 / 4.1 4.17

3 / 2.9 / 3 2.97

ANALYSING RESULTS AND CONCLUSION:

By looking at my results in the above tables, it has been found that as the force holding the two surfaces together (i.e. weight) increases the values for static and dynamic friction increases as well. It has also been found that the values for dynamic friction are less than the values for static friction.

I will only be analysing my average results. The reason being that my average results are probably more accurate as they have been found using numerous results. Therefore the following graphs show the average results for the dynamic/static friction for a particular surface and are plotted showing force against weight.

### Trends

The most common trends on my graphs are that as the weight increases so does its corresponding value for static and dynamic friction. It can also be assumed that the static and dynamic friction is directly proportional to the weight as all of my points lie either on or very near the line of best fit, they all show very strong positive correlations. I did some further research and found something called the coefficient of friction. This is a ratio and is defined as how much force in Newtons (N) needs to be applied to a weight of 1N in order for it to overcome friction on a particular surface, it is therefore a constant at all times. It can therefore be concluded that the gradient of a graph is equal to the coefficient of static/ dynamic friction of the particular surface it is representing. This can be seen in the following:

COEFFICIENT OF FRICTION = FRICTION (N)

WEIGHT (N)

GRADIENT = CHANGE IN Y AXIS

## CHANGE IN X AXIS

Since the Y-axis represents the force applied, which is equal to the frictional force according to Newton’s Second Law of Motion (see section entitled Background Information), and since the X-axis represents the total weight being exerted on the sand paper, the gradient is therefore equal to the coefficient of friction as long as the units for each axis is in Newtons (N).

The following are gradient calculations and the points I am using have been clearly marked on the graphs. However, since I have multiplied my results for the X-axis by 1000 in order to avoid dealing with small numbers when plotting my graphs I must now divide the results I am using for gradient calculations by 1000 in order for me to use my exact results.

For the graph showing the values found for static friction on the grade P60E sand paper surface:

The X-axis values for the points I’m using are 5500 and 2750 therefore to get the right numbers I must divide them by 1000. So, the values are actually 5.5 and 2.75

GRADIENT = CHANGE IN Y

CHANGE IN X

GRADIENT = 4.8 – 2.4

5.5 – 2.75

GRADIENT = 2.4

2.75

## GRADIENT = 0.873 (to 3 d.p)

So, this means that for every 1N of weight a force of 0.873N has to be applied to it in order for it to overcome static friction in the case of wood on grade P60E sand paper. Hence, the coefficient of static friction of wood on P60E sand paper is approximately 0.873 according to its definition.

For the graph showing the values found for dynamic friction on the grade P60E sand paper surface:

The X-axis values for the points I’m using are 6000 and 3000 therefore to get the right numbers I must divide them by 1000. So, the values are actually 6 and 3

GRADIENT = CHANGE IN Y

CHANGE IN X

GRADIENT = 4 - 2

6 – 3

GRADIENT = 2

3

GRADIENT = 0.667 (to 3 d.p)

So, this means that for every 1N of weight a force of 0.667N has to be applied to it in order for it to overcome dynamic friction in the case of wood on grade P60E sand paper. Hence, the coefficient of dynamic friction of wood on P60E sand paper is approximately 0.667 according to its definition.

For the graph showing the values found for static friction on the tabletop surface:

The X-axis values for the points I’m using are 8500 and 6000 therefore to get the right numbers I must divide them by 1000. So, the values are actually 8.5 and 6

GRADIENT = CHANGE IN Y

CHANGE IN X

GRADIENT = 4 – 2.8

8.5 – 6

GRADIENT = 1.2

2.5

GRADIENT = 0.48

So, this means that for every 1N of weight a force of 0.48N has to be applied to it in order for it to overcome static friction in the case of wood on the tabletop surface in my physics practical room. Hence, the coefficient of static friction of wood on the tabletop in question is approximately 0.48 according to its definition.

For the graph showing the values found for dynamic friction on the tabletop surface:

The X-axis values for the points I’m using are 6500 and 1050 therefore to get the right numbers I must divide them by 1000. So, the values are actually 6.5 and 1.05

GRADIENT = CHANGE IN Y

CHANGE IN X

GRADIENT = 2.4 – 0.4

6.5 – 1.05

GRADIENT = 2

5.45

GRADIENT = 0.367 (to 3 d.p)

So, this means that for every 1N of weight a force of 0.367N has to be applied to it in order for it to overcome dynamic friction in the case of wood on the tabletop surface in my physics practical room. Hence, the coefficient of dynamic friction of wood on the tabletop in question is approximately 0.367 according to its definition.

For the graph showing the values found for static friction on the grade P120 sand paper surface

The X-axis values for the points I’m using are 6500 and 1750 therefore to get the right numbers I must divide them by 1000. So, the values are actually 6.5 and 1.75

GRADIENT = CHANGE IN Y

CHANGE IN X

GRADIENT = 5.16 – 1.4

6.500 – 1.75

GRADIENT = 3.76

4.75

GRADIENT = 0.792 (to 3 d.p)

So, this means that for every 1N of weight a force of 0.792N has to be applied to it in order for it to overcome static friction in the case of wood on the grade P120 sand paper surface. Hence, the coefficient of static friction of wood on the grade P120 sand paper is approximately 0.792 according to its definition.

For the graph showing the values found for dynamic friction on the grade P120 sand paper surface

The X-axis values for the points I’m using are 5500 and 1500 therefore to get the right numbers I must divide them by 1000. So, the values are actually 5.5 and 1.5

GRADIENT = CHANGE IN Y

CHANGE IN X

GRADIENT = 3.6 – 1

5.5 – 1.5

GRADIENT = 2.6

4

GRADIENT = 0.65

So, this means that for every 1N of weight a force of 0.65N has to be applied to it in order for it to overcome dynamic friction in the case of wood on the grade P120 sand paper surface. Hence, the coefficient of dynamic friction of wood on the grade P120 sand paper is approximately 0.65 according to its definition.

Table 4: Table of Gradients

GRAPHS | GRADIENTS |

Graph 1: Static friction for P60E sand paper | 0.873 |

Graph 2: Dynamic friction for P60E sand paper | 0.667 |

Graph 3: Static friction for tabletop surface | 0.48 |

Graph 4: Dynamic friction for tabletop surface | 0.367 |

Graph 5: Static friction for P120 sand paper | 0.792 |

Graph 6: Dynamic friction for P120 sand paper | 0.65 |

Conclusion

Further Work

To improve the experiment in order to provide additional and more reliable evidence for a conclusion I would carry out the experiment as shown in the following diagram:

PULLEY

FORWARD FORCE

WOODEN BLOCK RESTING ON A HORIZONTAL

SURFACE

SCALE-PAN

This experiment doesn’t involve using a force meter; therefore, readings will be easier to take. Firstly, to find the value of static friction, the block’s mass is found by weighing it and then converted into its weight. Then, small masses are added, one at a time, to the scale-pan in order to increase the forward force. Eventually a point is reached at which the block starts to slide, and then the total mass that was placed in the scale-pan is converted into the total weight. This will be the value for the forward force and hence, by knowing this, the static frictional force can be found according to Newton’s Second Law of Motion which I have explained how to do in my section entitled BACKGROUND INFORMATION. Once this has been found, the coefficient of static friction can be found for wood on this particular surface as well. For dynamic friction, the experiment is carried out in almost the same way with the exception that the block is given a slight push each time a mass is added to the scale-pan. The value of the forward force at which the block continues to move with constant velocity after being pushed is the value of the dynamic frictional force. Once this has been found, the coefficient of dynamic friction can be found for wood on this particular surface as well.

This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.

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