Frictional Force Problem: Solve with 25kg & 0.36 Coefficient

[Josiah W]

The problem seems very straight forward, however it's proving stubborn. The problem is as follows: We have a mass of 25kg sliding on a table with a frictional coefficient of 0.36. What is the frictional force in Newtons? Now, I know the force of friction to be Fƒ = μmg and also Fƒ = μmg⋅cosα, but the angle in this case is 90 degrees, so I immediately ruled the second equation out. To me, this should be Fƒ = (0.36)(25kg)(9.81) = 88.29 Newtons of frictional force, but that apparently is incorrect. Any input on what I may be doing wrong, but may not understand about this process in general?

Thanks,
Josiah

 

[Bystander]

Significant figures?

 

[Big Red Car]

Hi Josiah, perhaps a diagram would make things easier to visualize:

I am not sure if cosα is required for the formula: Fƒ = μmg⋅cosα
My physics textbook "Physics for scientists and engineers with modern physics" by Serway/Jewett states that the friction force is obtained using the equations (pg 119-124; Chapter 5):

F_f = μ_s*(normal force); If given the static friction coefficient
F_f = μ_k*(normal force); if given the kinetic friction coefficient

The normal force is equal to mass*g if we know that there aren't any other forces acting in the vertical direction on our object.

You have stated the object is moving 'sliding' in the original problem. It's important to know if the 'sliding' force is purely horizontal or not.

So your first step is to figure out what the normal force is, and this will be dependent on whether the force applied to make the object slide has any vertical components.
When you know the normal force, you can then apply the relevant equations I have given above.

I believe this is how to do it, and hope my advice works out for you.

 

[Simon Bridge]

The problem seems very straight forward, however it's proving stubborn. The problem is as follows: We have a mass of 25kg sliding on a table with a frictional coefficient of 0.36. What is the frictional force in Newtons? Now, I know the force of friction to be Fƒ = μmg and also Fƒ = μmg⋅cosα, but the angle in this case is 90 degrees, ...

... do you mean to say that the table is vertical?

It looks like you are trying to guess the "correct" equation instead of understanding the physics.
The kinetic friction between two surfaces is ##\vec f = -\mu_kN\vec v/v## where ##\vec v## is the velocity of the sliding object, and ##\vec N## is the "normal force" - it acts perpendicular to the surface, and is usually what stops the object from falling through the surface.

This equation is saying that the friction has magnitude ##f=\mu_kN## and acts in the opposite direction to the motion.

You have to derive the equation for each specific case - which you do by drawing a free body diagram and using Newton's Laws.

 

[CWatters]

The problem seems very straight forward, however it's proving stubborn. The problem is as follows: We have a mass of 25kg sliding on a table with a frictional coefficient of 0.36. What is the frictional force in Newtons? Now, I know the force of friction to be Fƒ = μmg and also Fƒ = μmg⋅cosα, but the angle in this case is 90 degrees, so I immediately ruled the second equation out. To me, this should be Fƒ = (0.36)(25kg)(9.81) = 88.29 Newtons of frictional force, but that apparently is incorrect. Any input on what I may be doing wrong, but may not understand about this process in general?

I note that the problem statement says nothing about the velocity, acceleration or any other forces acting on it . Is that correct?

If so then I believe your answer is correct. Perhaps try g=10m/s/s rather than 9.81m/s/s ?

 

[Simon Bridge]

... or 9.8m/s/s ... if this is a computer marked thing, they can get very fussy about exact answers, to the correct number of decimal places.

 

[haruspex]

should be Fƒ = (0.36)(25kg)(9.81) = 88.29 N

As Bystander says, you are quoting too much precision for tne given inputs. If that is not the explanation, are you sure you have quoted the whole question word for word?