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Coefficient of static friction refers to the ratio between the maximum amount of static friction force (F) between the surfaces which are in close proximity to each other prior to the commencement of the movement to normal (N) force. It is generally termed as the friction force between two surfaces which are motionless.
Static Friction Definition
The tendency of relative motion while an item is at rest is known as static friction. If there is a normal force between the two surfaces and the object is slid, the force of static friction operates even before the object is slid.
Friction is a force that makes it more difficult for two items to slide past one another. The friction that exists between the surfaces on which an object is resting and a stationary object is known as static friction. Static friction, in other terms, is a force that holds an item at rest.
Kinetic Friction Definition
Kinetic friction is a force that opposes relative motion between two surfaces as they begin to move. The kinetic friction formula is as follows:
Fk= µ k N
- Coefficient of kinetic friction
Were,
μ k – Coefficient of friction
⇒ µ k = FkN
Coefficient of Static Friction Formula
F of Static friction = (coefficient of static friction) × (normal force)
Fs = µs x N
Where,
Fs is the force of static friction
µs is the coefficient of static friction and N is the Normal force
The coefficient of friction has no units because it is dimensionless. It's a scalar quantity, which means the force's direction has no bearing on the physical quantity. The coefficient of static friction is determined by the things that cause friction. Its value is usually between 0 and 1, but it can go higher. A number of 0 indicates that there is no friction between two items. When two items come into touch with each other, there will be some friction. The frictional force is equal to the normal force when the value is 1.
The coefficient of friction is frequently misunderstood to be limited to values between zero and one. A coefficient of static friction greater than one simply means that the frictional force is greater than the normal force. Silicone rubber, for example, can have a coefficient of static friction much above one.
Points to Remember based on Coefficient of Static Friction
- Coefficient of static friction is a part of unit 3, chapter 5 Laws of motion.
- It carries a total of 14 periods and 4 to 6 marks.
- Coefficient of static friction is the coefficient that signifies the relation between the normal force and frictional force between two bodies.
- It helps in understanding the amount of friction present between two surfaces of the bodies.
- Coefficient of static friction can be represented by μ = F/N
Sample Questions based on Coefficient of Static Friction
Ques 1 . A ten-kilogram object is placed on a smooth surface. The 15 N represents the static friction between these two surfaces. What is the static friction coefficient? (2 marks)
Ans: Given
10 kg = m, 30 N = F, µs =?
We are aware of this.
Normal force, N= mg
As a result, N = 10 x 9.81 = 98.1 N
The coefficient of static friction is calculated as follows:
F/N = µs
µs = 30/98.1
µs = 0.305
Ques 2. An object's normal force and static frictional force are 50 N and 80 N, respectively. What is the static friction coefficient? (1 mark)
Ans: Given
N = 50 N, F = 80 N, and µs =?
The coefficient of static friction is calculated using the following formula.
µs = F/N
µs = 80/50
µs = 1.6
Ques 3. A sled is subjected to a force of 5500 N. The sled's skis have a static friction coefficient of 0.75 with the snow. What is the maximum force of static friction if the fully loaded sled weighs 700 kg, and is the force provided sufficient to overcome it? Use the Formula for Coefficient of Static Friction. (1 mark)
Ans: The normal force on an object on a level surface is N= mg.
The force of static friction can be calculated using this method:
Fs= µs N
= µs mg
= 0.75 x 700 kg x 9.8 m/s2
= 5145 kgm/s2
= 5145 N
The maximum static friction force is 5145 N, so the applied force of 5500 N is sufficient to overcome it and start moving the sled.
Ques 4. What is the smallest static friction coefficient? (1 mark)
Ans: Only a limit value of zero can be used to determine the minimum coefficient of static friction. The value of the static friction coefficient must always be greater than zero; otherwise, the coefficient cannot be a friction coefficient.
Ques 5. How can you show that the static friction coefficient is "tangent" to the angle of repose? (1 mark)
Ans: It isn't tangential to. The tangent of the angle of repose is the static friction coefficient. Draw a figure based on the notion that two perpendicular angles, right side to right side and left side to left, are congruent.
Ques 6. How can I demonstrate that the coefficient of static friction is proportional to the mass of the thing in question, such as a wooden object? (2 marks)
Ans: The mass of the object has no bearing on the coefficient of friction. It is a distinguishing attribute of the combination of materials in touch, as well as its smoothness, molecular interaction, and lubrication ( and much more).
If you're wondering if friction is a factor, the answer is no. The friction force is calculated by multiplying the coefficient of friction by the normal (i.e., perpendicular) force exerted on the two bodies. Weight can (but does not have to) be included in the usual force.
Ques 7. What is the static friction coefficient between the crate and the ramp? (2 marks)
Ans: Static friction is a quality that may be measured rather than produced through modeling and analysis.
You might come upon a figure that someone has decided on, but it will only be that: a figure. You must do your tests, ideally in situ, if you truly require an estimate that is close to the true number.
If the estimate is incorrect, you must decide what judgments or safety problems may arise. If the possibilities of a problem are high and/or the severity is high, thorough testing is required. If you don't care, you can add a safety factor and run with it.
Ques 8. How do I calculate the frictional force between two blocks when one is on top of the other and the coefficients of static and dynamic friction are known (both are on a smooth surface)? (1 mark)
Ans: You have two blocks, one on top of the other, separated by a rough surface. Both blocks are also resting on a frictionless (smooth) surface. Because you didn't specify whether they were on an inclination or not, I'm going to presume the smooth surface is horizontal. Until an external force is applied to one (or both) of the blocks, there is no frictional force between them.
Ques 9. Is the normal force a factor in the coefficients of static and kinetic friction? (3 marks)
Ans: In physics problems, they are frequently assumed to be constants for specified pairs of materials. Friction, on the other hand, is essentially the locking together of electromagnetic forces. It goes to reason that if the normal force is great enough to change the surface's nature, the friction coefficients will change as well.
Ice skating is a good example of how this can be exploited to one's advantage. Sharp ice blades melt the ice locally and "smooth" it out, effectively lowering the coefficient of friction and making skating very energy efficient when compared to other methods of ice travel, such as blunt ice blades or boots with steel bottoms, which, despite being made of the same material as sharp ice blades, do not achieve the same level of ice surface alteration.
Ques 10. A 48kg crate is loaded onto a pick-up truck, and the truck speeds away without the crate sliding. If the coefficient of static friction between the truck and the crate is 0.4, what is the maximum acceleration that the truck can undergo without the crate slipping? (3 marks)
Ans: The truck must apply some sort of frictional force on the crate so that it does not slide. The frictional force is connected with the normal force through the coefficient of friction.
F = mg
Ffriction = μsFN = μsmg
This force of friction gets generated due to the acceleration of the truck on the basis of Newton’s second law.
FT = maT
When the truck is at its maximum acceleration without the moving of the crate the two forces will become equal.
FT = Ff
ma = μsmg
Acceleration:
a = μsg
a = (0.4)(9.8 m/s2)
A = 3.92 m/s2
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