# Coefficient of Friction Needed?

• haganjp
In summary, the problem involves an 80-kg skier sliding on waxed skis at constant velocity while pushing with poles. The horizontal component of the force pushing him forward is equal to the friction force, which is the product of the coefficient of friction and the normal force. However, since the coefficient of friction is not given, a solution cannot be determined. It is implied that the surface is frictionless, but the answer key states that 40N of force are required, which would require a coefficient of friction of 0.05. This information is not explicitly given in the problem.

## Homework Statement

I was reviewing some questions from the NYS Regents Physics exam and came across this question, which I could not answer. Here's the question ...

An 80-kg skier slides on waxed skis along a horizontal surface of snow at constant velocity while pushing with his poles. What is the horizontal component of the force pushing him forward?

## Homework Equations

In the horiztonal direction ...

Sum of Forces = mass x acceleration = 0 (constant velocity, therefore a=0)
Sum of Forces = Applied Force - Friction Force = 0

Applied Force = Friction Force
where, Friction Force = Coefficent of Friction x Normal Force

Normal Force = m x g = 80kg * 9.81 m/s^2 = 784.8 N

## The Attempt at a Solution

It seems I cannot determine a solution unless I know what the coefficient of friction is (which was not given in the problem statement). Am I missing something here?

It seems that it is implied that the surface is frictionless, in which case there is no need to apply a force to keep moving at a constant velocity, yet the answer key states that 40N of force are required. If this is indeed the answer, then a coefficient of friction of 0.05 is implied.

Thanks,

Jim

Yes, seems information given in question is insufficient.

Hi there,

From the information you gave in this problem, there is really an information missing. In my understanding also, the idea of the wax on the skis reduces the friction coefficient to a minimum.

Then again, $$\mu = 0.05$$ is not a very big coefficient, which could definitely be right.

For the Regents exam... somehow I'm not too surprised :-/

## 1. What is the coefficient of friction?

The coefficient of friction is a measure of the amount of resistance between two surfaces in contact. It represents the ratio of the force required to overcome the frictional force to the normal force between the two surfaces.

## 2. Why is the coefficient of friction important?

The coefficient of friction is important because it helps determine the amount of force needed to move an object across a surface. It also plays a crucial role in predicting the behavior of materials in various applications such as manufacturing, transportation, and sports.

## 3. How is the coefficient of friction measured?

The coefficient of friction can be measured by conducting experiments where the force needed to move an object across a surface is measured and divided by the normal force. This can also be calculated using mathematical models and simulations.

## 4. What factors affect the coefficient of friction?

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

## 5. Can the coefficient of friction be changed?

Yes, the coefficient of friction can be changed by altering the factors that affect it. For example, using a lubricant can reduce the coefficient of friction between two surfaces, making it easier to move an object across them.