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Solving Newton’s Second Law Problems Newton’s Second Law problems involve systems of one or more masses that are linked together and move as a unit. The simplest problems involve a single block sliding on a horizontal plane. Several complications can be added, such as friction, multiple blocks, inclined planes, and pulleys that change the direction of the system. All of these problems can be solved efficiently if you remember the basic premise: ΣF = ma The net force (or sum of forces) is what causes acceleration. There is no such thing as a “force of acceleration”. Again, Forces Cause Acceleration. Acceleration will always occur in the direction of the net force. A free body diagram (FBD) is the best way to show the forces on an object. You will use one or more FBDs to solve any second law problem. A FBD is simply a simplified diagram of an object on which you draw the force vectors that are applied to that object. For example, the following object has an outside force (F_{p}) and a friction force (F_{f}) applied to it, as well as the gravity and normal forces.
If it is obvious that some forces will cancel each other, they can be omitted, but when in doubt leave them in. In this case, F_{g} and F_{n} definitely cancel, but you will need F_{n} to calculate the value of F_{f} so it would be a good idea to leave them in. Look at the situation and figure out which way the object is likely to move. Call that the positive direction and draw an arrow near your FBD to remind you of which way is positive. Sometimes the positive direction isn’t quite clear, but take you best guess. If you’re wrong you can always go back and correct it later on. Once you get the FBD drawn, you are ready to write the second law equation. Start by writing the formal part of the equation: ΣF = ma Once this is done, add the forces that operate in the direction that the object will move. In this case, the object will move horizontally, so the only forces that concern us are horizontal: ΣF = ma = F_{f} + F_{p} You can plug in values for F_{f}, F_{p}, and m and solve for the acceleration. Of course, you could also write an equation for the vertical case: ΣF = ma = F_{g} + F_{n} Since the vertical acceleration is zero, F_{g} equals F_{n} in this case. Let’s say that you aren’t given a value for F_{f}. You can find it easily with the formula F_{f} = F_{n}μ_{k} F n is equal to F g as long as the block is on a horizontal surface, so you can say that F_{f} = mgμ_{k} (if the block is on an inclined surface, the friction formula is F_{f} = mgμ_{k}cosθ) Plugging the friction formula into the 2 nd law formula: ΣF = ma =  mgμ_{k} + F_{p} At this point, you can plug known values into the formula and solve for the unknowns. If you get a negative value for the acceleration, you have to make sure that it is correct. If the object is already moving in the positive direction, then a negative acceleration shows that the object is slowing down. If the object is stationary, a negative value for acceleration shows that the object will never move. You must be careful to make sure that the direction of your friction vectors oppose the motion (or attempted motion) of the object. Friction always opposes motion!
Atwood Machines The Atwood Machine was invented in 1784 by Rev. George Atwood as a laboratory experiment to verify the laws of uniformly accelerated motion. By slowing down the acceleration of the system, Atwood's machine allowed accelerated motion to be studied more easily in an era before photography and instrumentation were common. The ideal Atwood Machine consists of two masses, m1 and m2, connected by a massless cord over a massless pulley:
Once you get the positive direction straight, you should draw a FBD for each of the masses. This is easy because each mass only has two forces on it, a force due to gravity and an upward tension force.
Now you are ready to write the second law equation for this situation. Work your way through the system in the direction that it will likely move: ΣF = ma = F_{g2} + F_{T2}  F_{T2} + F_{g1} Remembering that F_{g1} = m_{1}g and F_{g2} = m_{2}g, and that the tension in a cord has to be the same at both ends, so ΣF = ma = m_{2}g + m_{1}g Remembering that m is the mass of the entire system, m_{1} + m_{2}, you can solve for the acceleration, a. The other thing you will be asked is the tension in the cord. Don’t make the mistake of thinking that it will simply be the weight of one of the masses, since this is a dynamic system and the masses are accelerating. The cord that supports the upwardmoving block has to account for both the weight of the block and for the force that accelerates the block upward. To solve for the tension in the cord all you have to do is isolate one of the masses and solve for the tension. For example, isolating block 2:
You can write the second law equation: ΣF = ma = F_{g2} + F_{T2} ΣF = ma = m_{2g} + F_{T2} The mass in this case will be just the mass of m2. You know a from the previous work, so you can solve for F_{T2}. You can do exactly the same thing for the other mass and the tension should come out the same. This is a great check for your work.
Inclined Planes Inclined planes present an interesting challenge to the solver. Unlike horizontal surfaces, gravity becomes an issue and has to be accounted for. Here is a typical inclined plane situation: The surface may or may not have friction. You make the FBD of the block like this: The gravity force is shown broken up into two components, F_{p} and F_{n}. Component F_{p} is parallel to the inclined plane and tends to cause the block to accelerate downhill. Component F_{n} is normal to the inclined plane. Force F_{n} is also the force that determines the friction force via the equation: F_{f} = F_{n}μ_{k} So, how do you determine all these forces? Well, F_{g} is easy, as we already know that F_{g} = mg For the other forces, use a little trig. Note that the angle at the top of the force triangle is the same as the angle of the incline. Since you know the hypotenuse of the force triangle, the other two sides are easily found: F_{p} = F_{g}sinθ F_{n} = F_{g}cosθ Once you know F_{n}, F_{f} is easily found with: F_{f} = F_{g}μ_{k}cosθ So, in a nutshell: F_{g} = mg F_{p} = F_{g}sinθ F_{n} = F_{g}cosθ F_{f} = F_{g}μ_{k}cosθ So, how do we use Newton’s second law to handle inclined plane problems? The only direction in which the object moves is parallel to the plane, so you should use only forces (or components of forces) that are parallel to the plane. The second law equation goes like this: ΣF = ma = F_{f} + F_{p} ΣF = ma = F_{g}μ_{k}cosθ + F_{g}sinθ ΣF = ma = mgμ_{k}cosθ + mgsinθ What if there are other forces involved like forces pulling the object up or down the hill? All you have to do is to find the parallel and normal components of those forces and include the parallel components in the second law equation. If a component of the outside force on the object is normal to the surface it will affect the normal force and you will have to calculate the new normal force to find the friction force.
Summary The basic steps in solving any Newton’s Second Law problem are:
