Coefficient of Friction, and frictional force

In summary: So F(net, x) = -F(friction) + F(pulling force, x) = 0, so F(friction) = F(pulling force, x).i think i understand now, so my initial attempt was correct, but the problem statement should have specified that static friction is applicable, instead of saying "if the coefficient of friction is 0.1" because as you just mentioned this is the maximum U, not the actual U. Thank you for the clarification. In summary, the first problem involves finding the coefficient of friction given the weight and frictional force of an object, which can be calculated using the equation F = uN. The second problem involves finding the frictional force of a block being pulled
  • #1
VladDracule
5
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Homework Statement


1st problem: If and object has a weight of 100 N and a frictional force of 50 N exerted on it, what is the coefficient of friction?

2nd problem: A Block with a mass of 50 kg, is pulled with a force of 10 N at an angle of 10 degrees to the horizon, if the coefficient of friction is 0.1, what is the frictional force?


Homework Equations


F = uN
with F = Frictional force
u = mew, coefficient of friction
N = normal force
Sin and Cos functions


The Attempt at a Solution



1) 50N = u*100N
50N/100N = u = 0.5
The solutions i found for this online says the answer is 0.2, which seems incorrect to me. Is my answer of 0.5 true?

2) F = (0.1) * [(50kg*9.8)-10sin(10)]
F = 48.8

the solution i found online to problem two uses the cos instead of sin function. This does not make any sense to me as the sin is what will give you the upwards force of the rope pulling the block right? Cos would give you the horizontal force which does not affect the normal, is my approach correct?
 
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  • #2
I am not sure for the first one as it seems there are other things that must be assumed that I am unsure of.

For the second, if the coefficient of friction is static friction (doesn't really matter if it is kinetic as static is usually larger) and the object is initially at rest (which should imply that it is static friction):

---
F_x = 10N * cos 10
F_y = 10N * sin 10

n = m*g - 10N * sin 10

F_s,max = 0.1 * (m*g - 10N * sin 10) = 0.1 * (50kg*9.8m/s^2 - 10N * sin 10) = 48.83N

10N * cos 10 = 9.848 N
---

What you must note is that mew of 0.1 is its max value. That is, if the force of 10N in the horizontal direction is less than mew,max * n, then mew * n = applied horizontal force.

So the frictional force should be 10N * cos 10 = ~9.85 N. This implies that we are taking some mew < 0.1 to make the net force zero.

If you used mew max, then your net force would be 9.85N - 48.84N = -38.99N, which implies that the object begins to accelerate in the opposite direction than the force which does not make sense. That is, if the object was initially at rest.
 
  • #3
I don't really follow what your saying on the second part. The general equation the solution that i found is using i s

F = 0.1 * {(m*g)-10cos(10)}
Which comes out to an answer of 48N
 
  • #4
VladDracule said:
I don't really follow what your saying on the second part. The general equation the solution that i found is using i s

F = 0.1 * {(m*g)-10cos(10)}
Which comes out to an answer of 48N
As sefrez noted, how can you get a frictional force in the negative x direction of 48 N if the pulling force in the x direction is only 10cos10 in the positive direction? The static friction force friction force is always less than uN unless the object is just on the verge of moving relative to the surface it is in contact with (in which case it is equal to uN, but that is not the case here). Use Newton'1st Law to find the friction force.
 
  • #5
Sefrez said:
So the frictional force should be 10N * cos 10 = ~9.85 N. This implies that we are taking some mew < 0.1 to make the net force zero.

If you used mew max, then your net force would be 9.85N - 48.84N = -38.99N, which implies that the object begins to accelerate in the opposite direction than the force which does not make sense. That is, if the object was initially at rest.

How does using cosine in the Force of friction equation even impact this in any way? that still doesn't make any sense because the cos tells you the horizontal force acting on the block, which as i said in my first question does not effect the normal force, which wouldn't affect the friction
 
  • #6
also the problem statement doesn't give any mention of it being static of kinetic friction
 
  • #7
VladDracule said:
also the problem statement doesn't give any mention of it being static of kinetic friction
Since the horizontal force (10 sin 10) is less than the maximum limiting value of the friction force, the block cannot move, and therfore, static friction applies, and static friction is less than or equal to uN. To find the static friction force, use Newton 1 in the x direction. It will not depend on N.
 

1. What is the definition of coefficient of friction?

The coefficient of friction is a measure of the amount of resistance between two surfaces in contact with each other. It is a dimensionless number that ranges from 0 to 1, with lower values indicating less friction and higher values indicating more friction.

2. How is coefficient of friction determined?

The coefficient of friction is determined by dividing the force required to move an object over a surface by the weight of the object. This can be done through experiments or by using known values for the materials in contact.

3. What factors affect the coefficient of friction?

The coefficient of friction can be affected by several factors, including the roughness of the surfaces, the materials in contact, the temperature, and the presence of any lubricants.

4. How is frictional force related to coefficient of friction?

The frictional force is directly proportional to the coefficient of friction. This means that as the coefficient of friction increases, the frictional force also increases. This relationship can be described by the equation F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force between the two surfaces.

5. How is coefficient of friction used in everyday life?

Coefficient of friction is used in many everyday situations, such as walking on different surfaces, driving a car, or using tools. It helps engineers design products with appropriate materials and surfaces to reduce friction and improve efficiency. It is also important in sports, as athletes often try to reduce friction to improve their performance.

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