Expert: Daniel Mazur - 11/26/2005

Question
Hi Daniel,
I hope I'm finding you in the best of health.
I'm in 11th grade and am 16 years old.
A 60kg boy is on a 15kg sled. He is pushed along a level path of snow where uk(coefficient of kinetic friction)=0.05.
What total force will accelerate it at 1.3 m/s/s?
Using an answer to a previous problem my teacher gave me as a paradigm, I think I know how to get the answer:
Ff=(.05)(75*9.81)=36.78
Fapplied=36.78+(75*1.3)
Fapplied=134N
Is this correct. If not, please explain why.
But what is wrong with this formula:
force of friction= m(-a).
Ff=(.05)(75)(-1.3).
When do I and when don't I use this above formula.
Here is an example of when my teacher used that formula:
A 10 kg mass is moving at 10m/s on a level surface. The kinetic coefficient of friction is .13. My teacher used that formula to find the acceleration in this problem.
Why can't i use this formula in the previous problem?
Thank you for your help.

Answer
Hi Jeff,

thank you for your question and in particular for the details about your education level.

Your first solution to the first problem is quite correct. Just for the sake of thorough understanding, please read on:

* You want to accelerate a 75kg object at 1.3 m/s^2, which requires force
F_0 = (75 * 1.3) N = 97.5 N, if no friction enters the problem.
* There is, however, a friction force to overcome here leading to an additional force, which you must ADD to the friction-less solution. The friction force is, in general,
Ff = -uk*m*g, in this case numerically
Ff = (-0.05 * 75 * 9.81) N = -36.8 N.
To OVERCOME the friction force, you have to ADD -Ff (yes, minus_Ff) to F_0. You see,
F_total = F_0 + (-fF), but as the Ff alone is NEGATIVE, (-Ff) means a POSITIVE contibution to the applied force.  The total force is then
F_total = F_0 + (-Ff) = 134 N.
The sign of the friction force is chosen negative to represent that friction's effect is always to DECELLERATE motion. When I use the expression "force to OVERCOME friction" above, I mean a force of the same magnitude and opposite sign to the friction force Ff. I use "friction compensation" below for the same thing.

The reason, why the alternative formula (based on m(-a)) will not work, lies in the handling of friction in calculus. Firstly, the friction compensation (-Ff) is ADDITIVE to the frictionless force F_0 (that is the solution of the same problem, when friction is neglected). In the alternative formula you gave, there is no trace of adding the two components F_0 and Ff. (You can say the F_0 is zero in this case - I quite agree, this reason of mine was given on the symbolic level, it's the second reason below that is crucial).

Secondly, the co-efficient of friction uk is not "just some number", you have to know, what you are allowed to do with it and what not. Quite generally, you apply the co-efficient to the respective >>pressure force between surfaces<< of two objects. In the very specific case of your problem this "pressure force" is identical with the weight of the massive object. Therefore, to get (-Ff) you multiply uk with this weight and, due to the previous paragraph reasoning, ADD the (-Ff) to your frictionless F_0.

Here's an example to broaden the picture: If the snow path was not level, the "pressure force" between the sledge and the path would NOT be the weight of the loaded sledge. It would be
F_press = (m*g)*cos(alpha), where (m*g) is the weight of the loaded sledge and alpha is the inclination of the slope. Your friction force
Ff = uk*F_press = uk*(m*g)*cos(alpha) (the parentheses are left just to stress out the origin of the term).
This paragraph is just an extension to your problem, please don't let it confuse you.

So to sum this part up, your alternative formula for Ff cannot work, because the "thing" you applied the co-efficient of friction uk to (the part ...(75)(-1.3)...) was NOT the "pressure force" between the two surfaces. The (75)(-1.3) = -97.5 N is a force, indeed, but it is the negative of the frictionless F_0! In the frictionless (!) case we would call this m*(-a) a "reactive force", i.e. the force with which the sledge pushes into your hands, while you are accelerating it.

Regarding the second problem, it shows, how useful it is to learn to do calculations with symbols first and plug numerical values in at the very end. Let me reproduce your teacher's derivation of the acceleration in the second problem here:

* The mass (M) is moving at a velocity (v_0) on a level surface. The only independent force in this problem is the weight (M*g) of the mass and it has two aspects - the tendency of the object to fall down and the friction (if there is anything at least partially compensating the downfall acceleration). Our friction is a positive value (uk), and the falling of the object is fully compensated by the level surface, so there is a friction force Ff=-uk*M*g and it it THE ONLY UNCOMPENSATED FORCE in the problem. Therefore,
F_total = Ff,
which means
M*a_total = -uk*M*g,
where a_total is the acceleration of the problem you want to  know. The only thing to do is to divide both sides of the equation by (non-zero) mass M and the result is

a_total = -uk*g.

* Only NOW one should consider plugging in the numbers, uk=0.13 and g=9.81 m/s^2, and the numerical result will be

a_total = -1.3 m/s^2.

* You may notice, that the problem contained redundant information about the object mass and velocity. I will leave to you the thinking, why the solution is independent of both the object mass and velocity. It is a good mind excercise ;-).

I hope the writing was clear enough to understand. To do any better, I'd need to draw a picture, which is, regretably, not possible at AllExperts.com.

Use it well and good luck in your science classes!
Daniel