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Lab report - coeffcient of friction

report on an experiment to determine the coefficient of friction
Course

General Physics I-Trig Based (PHY 201)

34 Documents
Students shared 34 documents in this course
Academic year: 2023/2024
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Jefferson Community College (New York)

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Coefficient of friction

Objective: To investigate the nature of the frictional forces between a smooth wooden block and a flat wooden board, in the horizontal and inclined plane.

Factors:

  1. ms Coefficient of static friction
  2. mk Coefficient of kinetic friction
  3. N Normal force of the block
  4. fs Frictional force (static)
  5. fk Frictional force (kinetic)
  6. m Mass
  7. g Gravity
  8.  Angle of the inclined plane
  9. T Tension of the string

Theory:

Friction is a resisting force that occurs when one object slides or attempts to slide over another. This force always acts at a tangent to the two surfaces, and in the opposite direction of the motion, or attempted motion. The friction depends upon the roughness of each surface, and the normal force, which is the force that the objects exert on each other. The normal force is always perpendicular to the surface.

The frictional force is directly proportional to the normal force, so the normal force (N) multiplied by a constant (μ) is equal to the frictional force. Note that the value for μ is dependent on the objects in contact, and if there is motion or not.

There are two types of friction, static friction (fs) and kinetic friction (fk), therefore we get the equations:

  1. fs = μsN

  2. fk = μkN

We put μs and μk because the value of μs is larger than μk, meaning the static frictional force is larger than the kinetic frictional force.

If we consider an inclined plane with an angle s, where s is the angle that the block begins to move, then we can find the normal force using:

  1. N = mgcoss

And we can find fs using:

  1. fs = mgsins

Combining these equations, we get:

  1. μs = = tans

Similarly, we can determine the normal force and fk by giving the block a slight push down the incline. The angle, k, should be recorded as the angle that the block slides down the incline with constant velocity, after the push.

  1. N = mgcosk and fk = mgsink

Again, combining the equations we get:

  1. μk = = tank

If we look at the horizontal plane, we have a mass attached to the block by a string, which passes over a pulley. Let the mass of the block by m1 and the added mass be m2. If we keep adding mass to m2, the point at which the block just moves is the point of maximum static friction. The conditions at that point are the following:

  1. T = fs T= m2g N = m1g fs = μsN

Combining these equations, we get:

  1. fs = m2g = μsN = μsm1g

Therefore:

m2g = μsm1g so,

  1. m2 = μsm

If we instead give the block a slight push, the point that it moves with constant velocity is the point of kinetic friction. The conditions at that point are the following:

  1. T = fk T = m2g N = m1g fk = μkN

Combining these equations, we get:

  1. m2 = μkm

Equations 10 and 11 mean that we can plot a graph of average m2 on the y-axis and m1 on the x-axis, with the gradient of this graph being the coefficient of either static or kinetic friction.

Apparatus:

 Wooden Block  Board  Small masses  Pulley  String  Scales

Sources of error: The equipment is very old so some parts of the block or board may have worn over time, and they are also contaminated with oils and other things. This means the friction may vary slightly over different points on the block and board. There also may have been human error when adding the masses to the mass holder, as it is difficult to place it without dropping it slightly (which will produce a momentarily larger force) or knocking the mass causing it to swing.

Data: Part 1 – Inclined Plane Static Friction

Trial 1 2 3 4 s 16 17 17 17.

Kinetic Friction Trial 1 2 3 4 k 12 12 12 12

Part 2 – Horizontal Plane Static Friction

Added Mass 0 kg 0 kg 0 kg 0 kg 0 kg m1 (kg) 0 0 0 0 1. m2 Trial 1 (kg) 0 0 0 0 0. m2 Trial 2 (kg) 0 0 0 0 0. m2 Trial 3 (kg) 0 0 0 0 0.

Kinetic Friction Added Mass 0 kg 0 kg 0 kg 0 kg 0 kg m1 0 0 0 0 1. m2 Trial 1 0 0 0 0 0. m2 Trial 2 0 0 0 0 0. m2 Trial 3 0 0 0 0 0.

Calculations Table:

Part 1 – Inclined Plane

Static Friction Kinetic Friction

Part 2 –

Horizontal Plane

Static Friction Kinetic Friction

Results:

Inclined Plane:

μs = 0.

μk = 0.

Horizontal Plane:

tans = μs Avg μs αμs

0 0 0.

tank = μk Avg μk αμk 0. 0 0 0. 0. 0.

m1 (kg) Avg m2 (kg) 0 0. 0 0. 0 0. 0 0. 1 0.

m1 (kg) Avg m2 (kg) 0 0. 0 0. 0 0. 0 0. 1 0.

It is clear from the results that μs > μk. This was apparent when determining them on a horizontal and inclined plane. However, when comparing values for μs on the inclined and horizontal plane, there is quite a large difference. On the inclined plane, the value for μs was 0 and on the horizontal plane the value for μs was 0. I believe the value of 0. is more accurate because there were less sources of error when performing the experiment on the inclined plane as it was a simple experiment. Despite this, the graph produced for μs on the horizontal plane had an r value of 0 which is very close to 1, meaning there was a linear relationship, but the larger value for μs was probably due to some systematic error.

When comparing the results for μk, they are much closer with values of 0 for the inclined plane and 0 for the horizontal plane. Again, I believe the value produced using the inclined plane was more accurate for the same reason as before, but the value produced on the horizontal plane was similar so this experiment was more accurate than that of calculating μs. The graph produced for μk in the horizontal plane also had a high r value (0) but a lower value than the graph for μs. Therefore, it is likely the horizontal plane experiment to calculate μk was more accurate than μs, but less precise, whereas the horizontal plane experiment to calculate μs was more precise but less accurate.

Further experimentation with better equipment would most likely yield better results. However, the results here still proved that μs > μk, and provided good results for both. To conclude, I feel that the experiment was a success and met the objectives, but could have been more accurate with better equipment and less human error.

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Lab report - coeffcient of friction

Course: General Physics I-Trig Based (PHY 201)

34 Documents
Students shared 34 documents in this course

University: Jefferson Community College (New York)

Was this document helpful?
Coefficient of friction
Objective: To investigate the nature of the frictional forces between a smooth wooden block
and a flat wooden board, in the horizontal and inclined plane.
Factors:
1. ms Coefficient of static friction
2. mk Coefficient of kinetic friction
3. N Normal force of the block
4. fs Frictional force (static)
5. fk Frictional force (kinetic)
6. m Mass
7. g Gravity
8. Angle of the inclined plane
9. T Tension of the string
Theory:
Friction is a resisting force that occurs when one object slides or attempts to slide over
another. This force always acts at a tangent to the two surfaces, and in the opposite
direction of the motion, or attempted motion. The friction depends upon the roughness of
each surface, and the normal force, which is the force that the objects exert on each other.
The normal force is always perpendicular to the surface.
The frictional force is directly proportional to the normal force, so the normal force (N)
multiplied by a constant (µ) is equal to the frictional force. Note that the value for µ is
dependent on the objects in contact, and if there is motion or not.
There are two types of friction, static friction (fs) and kinetic friction (fk), therefore we get
the equations:
1) fs = µsN
2) fk = µkN
We put µs and µk because the value of µs is larger than µk, meaning the static frictional
force is larger than the kinetic frictional force.
If we consider an inclined plane with an angle s, where s is the angle that the block begins
to move, then we can find the normal force using:
3) N = mgcoss
And we can find fs using:
4) fs = mgsins
Combining these equations, we get:

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