Physics force question plz save meh~?

Question

A box of books weighning 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 below the horizontal.

Find Mk(??? did i say it right? is it M or N) between the box and the floor.

Answer 1

I think you're looking for "mu - sub k" where that letter is actually the Greek letter mu, which in physics refers to a coefficient of friction. Any time an object moves over a surface, there is a resistance due to friction, and that is captured by the coefficient of friction. To find the resisting force due to friction, Ff, you need the normal force Fn and the coefficient of friction, mu:

Ff = mu * Fn

In this case, you have a force acting down on the box as a result of the 425 N, in addition to the weight of the box itself. The force resulting from the pushing can be found through some trig:

Fn1 = F * sin 35.2
Fn1 = (425 N)(sin 35.2)
Fn1 = 245 N

The force as a result of the weight of the box is 325 N. Thus, the total normal force Fn is the sum of the two, or 570 N.

We know that there is also a constant force acting parallel to the floor. Since this is a constant motion issue (the box isn't accelerating), the pushing force and the force of friction should be the same. So let's find that pushing force via more trig:

Fp = F * cos 35.2
Fp = (425 N)(cos 35.2)
Fp = 347 N

Now that we have all of our forces, we can find mu. Since our force of friction has to be the same as the pushing force, we can turn to our first equation:

Ff = mu * Fn
347 N = mu * 570 N
mu = 0.609

This is a unitless number and is always between 0 and 1.

By the way, there is also a mu - sub s, which is the coefficient of static friction. They are two different numbers - intuitively, that makes sense. It takes a little bit of effort to get a box moving, but once it's moving, it's not as hard to keep it going as it was to get it started.