Frictional Force Equation Doesn't Make Sense

[FredericChopin]

Frictional Force is mathematically defined as:

Ff = μ*m*g*cos(θ)

, where μ is the coefficient of friction, m is the mass of the object, g is the acceleration due to gravity and θ is the angle of the inclined plane.

But in terms of direction, this makes no sense!

Suppose there is an object on an inclined plane. It is going to travel down the plane because its Weight Force is pulling it towards the Earth. At the same time, a Frictional Force will act in the opposite direction to the motion of the object.

The Frictional Force, however, is calculated using the object's Normal Force times cos(θ) (times μ), which acts perpendicular to the surface of the plane. But this means that the Normal Force is not acting in the same axis as the Frictional Force, so how can Normal Force times cos(θ) (times μ) be used to calculate Frictional Force?

Let me elaborate: In order to calculate Normal Force in the same axis as Frictional Force, you should use μ*m*g*sin(θ), not μ*m*g*cos(θ). Would it not make more sense this way?

I'm either very frustrated or very confused. Maybe a diagram with your answer can help.

Thank you.

 

[jedishrfu]

But it makes sense if you think that you're breaking up the gravitational force into two components one parallel to the motion and one perpendicular to the motion. Its the perpendicular component magnitude that dictates the friction not its direction.

It makes sense if you consider motion on a horizontal plane vs motion on an incline of say 45 degree vs motion sliding down a vertical plane. In the first case its the full weight of the object(cos(0)*weight vs cos(45)*weight vs cos(90)* weight (or zero frictional force).

 

[WannabeNewton]

The magnitude of the frictional force can be related to the magnitude of the normal force by $f\leq \mu N$. When slipping occurs, we have $f= \mu N$. If a block is sliding down an inclined plane of angle $\theta$ (i.e. slipping relative to the surface of the inclined plane), Newton's 2nd law gives for the motion along the plane and perpendicular to the plane, respectively, $mg\sin\theta - f = ma$ and $mg\cos\theta - N =0$ (I have set up my coordinates so that the x-axis is along the incline and the y-axis is perpendicular to the incline). We can then easily find an expression for the magnitude of friction $f$ and we find that $f = \mu mg\cos\theta$. In summary, we are relating the magnitudes of the normal force to the frictional force, not their directions! The direction of $f$ is of course along the incline.

 

[FredericChopin]

A New Question

Thank you jedishrfu and WannabeNewton; that cleared a lot of things up. But this leaves a different question.

Why is Normal Force proportional to Frictional Force? In other words:

"...we find that f=μmgcosθ...".

, but why?

I think this would require more of an explanation than an equation.

 

[WannabeNewton]

I believe you are asking for a justification of why $f\leq \mu N$ holds true for sliding friction. Keep in mind that it is an approximate model of sliding friction and is known as Coulomb friction; it is justified via experiment. http://en.wikipedia.org/wiki/Friction#Laws_of_dry_friction

 

[jedishrfu]

I believe you are asking for a justification of why $f\leq \mu N$ holds true for sliding friction. Keep in mind that it is an approximate model of sliding friction and is known as Coulomb friction; it is justified via experiment. http://en.wikipedia.org/wiki/Friction#Laws_of_dry_friction

Good point, there is also an initial frictional force that must be overcome to start something sliding that is higher than the sliding frictional force.

 

[FredericChopin]

Brilliant. Thank you guys.

 

[SammyS]

Thank you jedishrfu and WannabeNewton; that cleared a lot of things up. But this leaves a different question.

Why is Normal Force proportional to Frictional Force? In other words:

"...we find that f=μmgcosθ...".

, but why?

I think this would require more of an explanation than an equation.

Considering "Cause and Effect", I would express that as the magnitude of the Frictional Force is directly proportional to the magnitude of the Normal Force. The resulting equation is the same, but in my view it's the amount of the Normal Force which produces the force of friction.

 

[Chestermiller]

Frictional Force is mathematically defined as:

Ff = μ*m*g*cos(θ)

, where μ is the coefficient of friction, m is the mass of the object, g is the acceleration due to gravity and θ is the angle of the inclined plane.

But in terms of direction, this makes no sense!

Suppose there is an object on an inclined plane. It is going to travel down the plane because its Weight Force is pulling it towards the Earth. At the same time, a Frictional Force will act in the opposite direction to the motion of the object.

The Frictional Force, however, is calculated using the object's Normal Force times cos(θ) (times μ), which acts perpendicular to the surface of the plane. But this means that the Normal Force is not acting in the same axis as the Frictional Force, so how can Normal Force times cos(θ) (times μ) be used to calculate Frictional Force?

Let me elaborate: In order to calculate Normal Force in the same axis as Frictional Force, you should use μ*m*g*sin(θ), not μ*m*g*cos(θ). Would it not make more sense this way?

I'm either very frustrated or very confused. Maybe a diagram with your answer can help.

Thank you.

You are confused because μ*m*g*cos(θ) is not the frictional force. It represents the maximum frictional force that can be sustained without the block beginning to slide down the plane. The block will begin to slide when the angle is just large enough for mg sin(θ) to equal μ*m*g*cos(θ). Otherwise, the frictional force will be less than μ*m*g*cos(θ), and the block will not slide.