Application of Newton's Second Law
When applying the second law, a very important first step is to establish a coordinate system. This is necessary because the second law requires us to add up force vectors and equate them such that or and this is always done by adding up vectors by components. Components are, by definition, the projections of a vector along the coordinate axes. Without first establishing a coordinate system, we cannot begin to find these components.
How to Choose a Coordinate System
The choice of coordinate system can often make the difference between a problem being relatively easy to solve and being nearly impossible to solve. A general rule in choosing the best coordinate system is that you want one of the coordinate axes to line up with the direction of acceleration of the system, if possible. This is so important that we sometimes choose to deal with polar coordinates in vector form (as we'll do for circular motion) rather than breaking this rule and sticking with rectangular (also called Cartesian) coordinates.
For instance, given a skateboarder going down a hill, you might be inclined (no pun intended) to use the common choice of the x-axis pointing in the horizontal direction and the y-axis pointing upward. If you try solving the problem this way it is very difficult. A much better choice is to make the x-axis point parallel to the hill, so that the skateboarder will travel along and accelerate along this axis. Solving the problem this way is much easier.
As another example, projectile motion without air drag is best solved using the standard +y upward and +x horizontal coordinate system, but only because the acceleration is parallel to the -y axis and not for some other reason.
We will soon see that circular motion is best solved using a polar coordinate system such that the radial direction is parallel to the acceleration. If the acceleration varies in a complicated fashion, chances are the problem will need to be solved using numerical methods and won't be one you can solve by hand.
Finding Forces
Here are some guidelines for how to proceed after the coordinate system is established. They won't apply to every case, but certainly will work for many cases:
- You should look at any known forces and express them in the chosen coordinates. For instance, the force of gravity is a known vector if the problem relates to an object near earth's surface. The downward force of gravity may need to be broken into components if one of your coordinate axes is not directed downward, as in the case of the skateboarder mentioned above.
- Once you know the gravitational force, you can usually determine the normal force that a surface (if present) will exert on the object.
- Once the normal force (if present) is known you will be able to express the force of friction, if the surface isn't frictionless.
Block on a Level Surface
When an object rests on a level surface (assumed on earth), the forces in the vertical direction will always add to zero. One force of which you are guaranteed is the downward gravitational force on the object, Besides this, there will typically be a normal force The only way the normal force will not be present is if some other force besides the surface is supporting all of the weight of the object. This may occur, for instance, if there is some other applied force on the object that is caused by some person who is pulling on the object.
In the absence of such an applied force, we simply find that the normal force on the block due to the surface on which it rests becomes just big enough (by compressing molecular springs) to oppose the gravitational force downward on the block due to the planet's invisible tug of gravity.
If an applied force is present, and if it has a vertical component, then the normal force will change. For instance, if the applied force pulls a bit upward, then the normal force doesn't have to support as much of the block's weight, and therefore decreases by an amount equal to the vertical pull of the applied force.
Another way to see this is that so long as the block stays on the ground, the vertical components of all the forces need to add to zero. Recall that we will always assume our surface is strong enough to support the block, so accelerating downward will not be an option. Accelerating upward, however, is possible. If the upward component of the applied force ever exceeds the gravitational force downward, the object will accelerate into the sky! Certainly no normal force exists in such a scenario.
Mathematically, so long as the object is on the ground we may write: Recall that in the subscripts that 'g' is for gravity, 'N' is for the normal force and 'A' is an applied force. Be careful to note that those terms are the y-components of the force vectors, and may be either positive or negative valued (or zero). For instance, if we take the +y direction to be upward. The normal force may only ever be upward, and never pulls objects downward toward the ground. The applied force may have either a positive or negative vertical component depending on the way the force is directed.
If we look at the horizontal direction, the only two possible forces on a level surface are the x-component of the applied force, and the opposing force of friction, if there is any. So without friction is the only horizontal force. With friction there will be an opposing force that may be as small as zero (if the applied force isn't trying to slide the box) and as large as
Looking at this, you can see that since the applied force in the vertical direction affects the normal force, that it will also have an effect on the horizontal friction force. So mathematically we can write:
Just as before, the terms may be positive or negative. If we call the rightward direction positive, then if the applied force switches direction, it will be negative. Be careful to note that the friction force will in this static (stationary) situation always oppose the direction of the applied force's horizontal component. If the applied force pulls rightward, friction goes leftward. If the applied force goes leftward, the friction goes rightward.
All of this discussion is animated in the interactive graphic below. Play around with changing the applied force vector and see how the other vectors respond.
Forces on a block resting on a horizontal surface
A Numerical Example
Suppose the block in the graphic above has a mass of 5 kg and has a friction coefficient of 0.40. Also suppose the applied force is 20N directed at 30 degrees above the horizontal. Given these values, let's compute all the forces. The gravitational force is downward and mg=5kg(10m/s2)=50N.
Before we can determine the value of the normal force, we need to know the upward pull of the applied force. It is 20Nsin(30o)=10N. Therefore we can write the sum of vertical forces as FAy+FNy+Fgy=0. Plugging in gives 10N+FNy-50N=0. So FNy=40N. Note that there is no x-component on the normal force or the gravitational force, so their y-components are the whole vectors.
Now that we know the normal force, the friction force may be determined from the friction coefficient and the normal force: Ff=0.40(40N)=16N. Given the rightward pull of the applied force equal to FAx=20Ncos(30o)=17.3N, the net force in the x-direction is 17.3N-16N = 1.3N... or just a little force to start the block sliding. Keep in mind that the moment after the sliding begins, the friction coefficient would typically become a bit smaller. This would mean the friction force would decrease and the net force to the right would rise a bit.
A More Sophisticated Example
Based on our limited discussions about a block on a level surface with friction, we can ask an interesting question: If I wish to pull a load as quickly as possible with a given force, at what angle should I pull? I hope you noticed in the math above that the applied force affects the value of the friction force. The larger the vertical component of the applied force, the smaller the horizontal friction force becomes. If we wanted to pull the block as quickly as possible to the right, it'd seem that pulling upward would help since friction would decrease. The other factor, however, is that as we pull in a more upward direction, less of our effort is directed rightward - which is the whole point of the pull. So it seems there is perhaps an optimal angle to pull at such that the rightward component of the pull is still relatively large, but the upward component does a decent job of reducing the friction via a reduction of the normal force. It turns out there is an optimal angle. Let's find it.
If we assume the goal is to drag the block as quickly as possible, what we desire is that As before, we get the normal force in the y-direction using Don't forget here that g is not negative-valued! It is the magnitude of a vector which points downward. Those are not the same thing.
Knowing the normal force allows us to find the friction force: The other horizontal force is just the x-component of the applied force, or So these two forces need to lead to a maximum horizontal force by choice of the best value of theta, or:
We are looking for a maximum value based on choice of theta, which means we need to take the derivative of this expression with respect to theta and set it equal to zero. This is because we are always guaranteed a maximum or minimum when the derivative is zero. It'd be a good idea when we get the result to verify that we do indeed get a maximum and not a minimum. The derivative is: Solving the algebra leads to The interpretation of this is as follows: The optimum angle to pull the block depends on the friction coefficient. If we know the value of the friction coefficient, then the ideal angle to pull at is the arc tangent of that coefficient. If you think back to the tangent function (or look below), you'll see that it implies that if there is no friction the best idea is to pull directly rightward (theta=0), which I hope matches your intuition. As friction becomes larger, the optimal angle grows as well. However, regardless of how large the friction coefficient becomes, the appropriate angle will never exceed 90o, but will only approach it asymptotically as the tangent function does.