Physics IX: Force Problems

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Physics IX: Force Problems

Author: oLahav

This series of lessons is designed to help you learn, or review, the fundamentals of physics. This part will show you how to solve force problems.

Force problems? Sounds dangerous!

Have no fear, force problems are actually quite easy. There are a few distinct types, but they all share the same solving mechanism. So, without further ado, here's what you do when you're trying to solve a force problem:

Draw an FBD. Remember to decide on which directions are positive and which are negative, note all forces acting on the relevant object, and the motion of the object.
Fill in what you know- the values of the forces you're given, and what you cna calculate.
Use vector addition to add up the various forces and find net force.
Use net force to find other values, like acceleration, if you have to.

Sounds simple, and it is, but it takes some practice. Let's look at a couple of examples:

Example 1

"An object of 5 kg is resting of a slope that is 45 degrees off the ground. What is the coefficient of static friction between the object and the slope?

So, let's start by drawing an FBD:

Good. now, we know that F_N=mu* F_s , and we're looking for find mu. We know that there is no motion, and that this means the net force

by Newton's First Law.

Before we go any further, let's determine directions in our picture. We can call "up the slope" our positive x direction, and "perpendicular to the slope" would be our y direction, so we rotate the picture like so:

Now, we have a diagonal force of gravity, so let's split it into 2- the horizontal component (which equals the friction) and the vertical (which equals the normal). We know that F_g=9.8m=9.8(5)=49 N , and the degree we have there is 45, so F_N=49 * sin(45)=34.6 N and F_s=49 * cos(45)=34.6 N . Now all we do is divide the two to find that mu=1 .

See? This was pretty simple. The trick of rotating the axes is useful (it does work perfectly well), remember that.

Example 2

A box of 20 kg is pushed across a surface and then released with an initial speed of 10 m/s. What will the object's speed be after 1 second if the coefficient of friction is 0.5?

Again, FBD time:

Now, we don't even need to invert anything. Note the motion is opposite to friction, gravity and normal force cancel out, etc. We're looking for speed, so it may seem like we don't need forces at all, right? But, as you'll see, forces will give us acceleration, which we need for speed (hey, cool, NfS).

Anyway, you may ask yourself, what happened to the push? Shouldn't that go into the FBD as applied force? Well, no. Note that we're not pushing the object any longer, so applied force isn't working on the box now.

So, let's calculate our net force, which will really just be friction. F_f=muF_N , remember? And F_N=F_g=9.8m=9,8(20)=196N , so F=(0.5)(196)=98 N . But, remember that the force is acting in a reverse direction to motion, so F=-98 N

Now, remember from Newton's Second Law that F=ma , so $a=\frac{F}{m}=\frac{-98}{20}=-4.9 m/s^2$ .

And, from way back when, $a=\frac{v_2-v_1}{t}$ , so v_2=at+v_1=-4.9(1)+10=5.1 m/s . So, after all that, our speed after 1 second reduced to only 5.1 m/s. Cool!