In this lesson, students will learn how to solve difficult problems using Newton's 2nd law.
Key Equations
\begin{align*}\mathrm{Common~Forces} \begin{cases} F_g = m g & \text{Gravity}\\ F_N & \text{Normal force: acts perpendicular to surfaces}\\ F_T & \text{Force of tension in strings and wires}\\ F_{sp}= k \Delta x = & \text{Force of spring} \Delta x \text{from equilibrium}\end{cases}\end{align*}
\begin{align*}\text{Force Sums} \begin{cases} F_{\text{net}} = \sum_{i} F_i = m a & \text{Net force is the vector sum of all the forces}\\ F_{\text{net,}x} = \sum_{i} {F_{ix}} = m{a}_{x} & \text{Horizontal components add also}\\ F_{\text{net,}y} = \sum_{i} {F_{iy}} = m{a}_{y} & \text{As do vertical ones} \end{cases}\end{align*}
\begin{align*}\text{Static and Kinetic Friction'} \begin{cases} f_s \le \mu_s  F_N & \text{Opposes potential motion of surfaces in contact}\\ f_k = \mu_k  F_N & \text{Opposes motion of surfaces in contact} \end{cases}\end{align*}
Ultimately, many of these “contact” forces are due to attractive and repulsive electromagnetic forces between atoms in materials.
Guidance
Problem Solving for Newton’s Laws, StepByStep

Figure out which object is “of interest.”
 If you're looking for the motion of a rolling cart, the cart is the object of interest.
 If the object of interest is not moving, that's OK, don't panic yet.
 Draw a sketch! This may help you sort out which object is which in your problem.

Identify all the forces acting on the object and draw them on object. (This is a freebody diagram FBD)
 If the object has mass and is near the Earth, the easiest (and therefore, first) force to write down is the force of gravity, pointing downward, with value mg.
 If the object is in contact with a flat surface, it means there is a normal force acting on the object. This normal force points away from and is perpendicular to the surface.
 There may be more than one normal force acting on an object. For instance, if you have a bologna sandwich, remember that the slice of bologna feels normal forces from both the slices of bread!
 If a rope, wire, or cord is pulling on the object in question, you've found yourself a tension force. The direction of this force is in the same direction that the rope is pulling.
 Don't worry about any forces acting on other objects. For instance, if you have a bologna sandwich as your object of interest, and you're thinking about the forces acting on the slice of bologna, don't worry about the force of gravity acting on either piece of bread.
 Remember that Newton's \begin{align*}3^{rd}\end{align*} Law, calling for “equal and opposite forces,” does not apply to a single object. None of your forces should be “equal and opposite” on the same object in the sense of Newton's \begin{align*}3^{rd}\end{align*} Law. Third law pairs act on two different objects.
 Recall that scales (like a bathroom scale you weigh yourself on) read out the normal force acting on you, not your weight. If you are at rest on the scale, the normal force equals your weight. If you are accelerating up or down, the normal force had better be higher or lower than your weight, or you won’t have an unbalanced force to accelerate you.
 Never include “ma” as a force acting on an object. “ma” is the result of the net force \begin{align*}F_{net}\end{align*} which is found by summing all the forces acting on your object of interest.

Determine how to orient your axes
 A good rule to generally follow is that you want one axis (usually the xaxis) to be parallel to the surface your object of interest is sitting on.
 If your object is on a ramp, tilt your axes so that the xaxis is parallel to the incline and the yaxis is perpendicular. In this case, this will force you to break the force of gravity on the object into its components. But by tilting your axes, you will generally have to break up fewer vectors, making the whole problem simpler.

Identify which forces are in the \begin{align*} x\end{align*} direction, which are in the \begin{align*}y\end{align*} direction, and which are at an angle.
 If a force is upward, make it in the \begin{align*}y\end{align*}direction and give it a positive sign. If it is downward, make it in the \begin{align*}y\end{align*}direction and give it a negative sign.
 Same thing applies for right vs. left in the \begin{align*}x\end{align*}direction. Make rightward forces positive.
 If forces are at an angle, draw them at an angle. A great example is that when a dog on a leash runs ahead, pulling you along, it’s pulling both forward and down on your hand.
 Draw the free body diagram (FBD).
 Remember that the FBD is supposed to be helping you with your problem. For instance, if you forget a force, it'll be really obvious on your FBD.
 Break the forces that are at angles into their \begin{align*}x\end{align*} and \begin{align*}y\end{align*} components
 Use right triangle trigonometry
 Remember that these components aren't new forces, but are just what makes up the forces you've already identified.
 Consider making a second FBD to do this component work, so that your first FBD doesn't get too messy.

Add up all the \begin{align*}x\end{align*}forces and \begin{align*}x\end{align*}components.
 Remember that all the rightward forces add with a plus \begin{align*}(+)\end{align*} sign, and that all the leftward forces add with a minus \begin{align*}()\end{align*} sign.
 Don't forget about the \begin{align*}x\end{align*}components of any forces that are at an angle!
 When you've added them all up, call this "the sum of all \begin{align*}x\end{align*} forces" or "the net force in the \begin{align*}x\end{align*}direction."

Add up all the \begin{align*}y\end{align*}forces and \begin{align*}y\end{align*}components.
 Remember that all the upward forces add with a \begin{align*}(+)\end{align*} sign, all the downward forces add with a \begin{align*}()\end{align*} sign.
 Don't forget about the \begin{align*}y\end{align*}components of any forces that are at an angle!
 When you've added them all up, call this "the sum of all \begin{align*}y\end{align*} forces" or "net force in the \begin{align*}y\end{align*}direction."

Use Newton's Laws twice.
 The sum of all \begin{align*}x\end{align*}forces, divided by the mass, is the object's acceleration in the \begin{align*}x\end{align*}direction.
 The sum of all \begin{align*}y\end{align*}forces, divided by the mass, is the object's acceleration in the \begin{align*}y\end{align*}direction.
 If you happen to know that the acceleration in the \begin{align*}x\end{align*}direction or \begin{align*}y\end{align*}direction is zero (say the object is just sitting on a table), then you can plug this in to Newton’s \begin{align*}2^{nd}\end{align*} Law directly.
 If you happen to know the acceleration, you can plug this in directly too.

Each body should have a FBD.
 Draw a separate FBD for each body.
 Set up a sum of forces equation based on the FBD for each body.
 Newton’s Third Law will tell you which forces on different bodies are the same in magnitude.
 Your equations should equal your unknown variables at this point.
Example 1
Question: Using the diagram below, find the net force on the block. The block weighs \begin{align*}3\mathrm{kg}\end{align*} and the inclined plane has a coefficient of friction of \begin{align*}0.6\end{align*}.
Answer:
The first step to solving a Newton's Laws problem is to identify the object in question. In our case, the block on the slope is the object of interest.
Next, we need to draw a freebody diagram. To do this, we need to identify all of the forces acting on the block and their direction. The forces are friction, which acts in the negative x direction, the normal force, which acts in the positive y direction, and gravity, which acts in a combination of the negative y direction and the positive x direction. Notice that we have rotated the picture so that the majority of the forces acting on the block are along the y or x axis. This does not change the answer to the problem because the direction of the forces is still the same relative to each other. When we have determined our answer, we can simply rotate it back to the original position.
Now we need to break down gravity (the only force not along one of the axises) into its component vectors (which do follow the axises). \begin{align*} \mathrm{The~x~component~of~gravity:}~9.8\mathrm{m/{s^2}}\times cos{60}=4.9\mathrm{m/{s^2}}\\ \mathrm{The~y~component~of~gravity:}~9.8\mathrm{m/{s^2}}\times sin{60}=8.5\mathrm{m/{s^2}} \end{align*} Yet these are only the acceleration of gravity so we need to multiply them by the weight of the block to get the force. \begin{align*} F=ma=3\mathrm{kg}\times 4.9\mathrm{m/{s^2}}=14.7\mathrm{N} F=ma=3\mathrm{kg}\times 8.5\mathrm{m/{s^2}}=25.5\mathrm{N} \end{align*}
Now that we have solved for the force of the ycomponent of gravity we know the normal force (they are equal). Therefore the normal force is \begin{align*}25.5 \mathrm{N}\end{align*}. Now that we have the normal force and the coefficient of static friction, we can find the force of friction. \begin{align*} F_s=\mu _sF_N=.6\times 25.5\mathrm{N}=15.3\mathrm{N} \end{align*}
The force of static friction is greater than the component of gravity that is forcing the block down the inclined plane. Therefore the force of friction will match the force of the xcomponent of gravity. So the net force on the block is \begin{align*} \mathrm{net~force~in~the~xdirection:}~\overbrace{14.7\mathrm{N}}^{xcomponent~of~gravity}\overbrace{14.7\mathrm{N}}^{force~of~friction}=0\mathrm{N}\\ \mathrm{net~force~in~the~ydirection:}~\underbrace{25.5\mathrm{N}}_{Normal~Force}\underbrace{25.5\mathrm{N}}_{ycomponent~of~gravity}=0\mathrm{N} \end{align*} Therefore the net force on the block is \begin{align*}0\mathrm{N}\end{align*}.
Example 2
Watch this Explanation
Simulation
Explore More
 Find the mass of the painting. The tension in the leftmost rope is \begin{align*}7.2\;\mathrm{N}\end{align*}, in the middle rope it is \begin{align*}16 \;\mathrm{N}\end{align*}, and in the rightmost rope it is \begin{align*}16 \;\mathrm{N}\end{align*}.
 Find Brittany’s acceleration down the frictionless waterslide in terms of her mass \begin{align*}m\end{align*}, the angle \begin{align*}\theta\end{align*} of the incline, and the acceleration of gravity \begin{align*}g\end{align*}.
 The physics professor holds an eraser up against a wall by pushing it directly against the wall with a completely horizontal force of \begin{align*}20 \;\mathrm{N}\end{align*}. The eraser has a mass of \begin{align*}0.5 \;\mathrm{kg}\end{align*}. The wall has coefficients of friction \begin{align*}\mu_S = 0.8\end{align*} and \begin{align*}\mu_K = 0.6.\end{align*}
 Draw a free body diagram for the eraser.
 What is the normal force \begin{align*}F_N\end{align*} acting on the eraser?
 What is the frictional force \begin{align*}F_S\end{align*} equal to?
 What is the maximum mass m the eraser could have and still not fall down?
 What would happen if the wall and eraser were both frictionless?
 A tractor of mass \begin{align*}580 \;\mathrm{kg}\end{align*} accelerates up a \begin{align*}10^\circ\end{align*} incline from rest to a speed of \begin{align*}10 \;\mathrm{m/s}\end{align*} in \begin{align*}4 \;\mathrm{s}\end{align*}. For all of answers below, provide a magnitude and a direction.
 What net force \begin{align*}F_{net}\end{align*} has been applied to the tractor?
 What is the normal force, \begin{align*}F_N\end{align*} on the tractor?
 What is the force of gravity \begin{align*}F_g\end{align*} on the tractor?
 What force has been applied to the tractor so that it moves uphill?
 What is the source of this force?
 A heavy box (mass \begin{align*}25 \;\mathrm{kg}\end{align*}) is dragged along the floor by a kid at a \begin{align*} 30^\circ\end{align*} angle to the horizontal with a force of \begin{align*}80 \;\mathrm{N}\end{align*} (which is the maximum force the kid can apply).
 Draw the free body diagram.
 What is the normal force \begin{align*}F_N\end{align*}?
 Does the normal force decrease or increase as the angle of pull increases? Explain.
 Assuming no friction, what is the acceleration of the box?
 Assuming it begins at rest, what is its speed after ten seconds?
 Is it possible for the kid to lift the box by pulling straight up on the rope?
 In the absence of friction, what is the net force in the \begin{align*}x\end{align*}direction if the kid pulls at a \begin{align*}30^\circ\end{align*} angle?
 In the absence of friction, what is the net force in the \begin{align*}x\end{align*}direction if the kid pulls at a \begin{align*}45^\circ\end{align*} angle?
 In the absence of friction, what is the net force in the \begin{align*}x\end{align*}direction if the kid pulls at a\begin{align*} 60^\circ\end{align*} angle?
 The kid pulls the box at constant velocity at an angle of \begin{align*}30^\circ\end{align*}. What is the coefficient of kinetic friction \begin{align*}\mu_K\end{align*} between the box and the floor?
 The kid pulls the box at an angle of \begin{align*}30^\circ\end{align*}, producing an acceleration of \begin{align*}2 \;\mathrm{m/s}^2\end{align*}. What is the coefficient of kinetic friction \begin{align*}\mu_{K}\end{align*} between the box and the floor?
 Spinal implant problem — this is a real life biomed engineering problem! Here’s the situation: both springs are compressed by an amount \begin{align*}x_o\end{align*}. The rod of length \begin{align*}L\end{align*} is fixed to both the top plate and the bottom plate. The two springs, each with spring constant \begin{align*}k\end{align*}, are wrapped around the rod on both sides of the middle plate, but are free to move because they are not attached to the rod or the plates. The middle plate has negligible mass, and is constrained in its motion by the compression forces of the top and bottom springs. The medical implementation of this device is to screw the top plate to one vertebrae and the middle plate to the vertebrae directly below. The bottom plate is suspended in space. Instead of fusing broken vertebrates together, this implant allows movement somewhat analogous to the natural movement of functioning vertebrae. Below you will do the exact calculations that an engineer did to get this device patented and available for use at hospitals.
 Find the force, \begin{align*}F\end{align*}, on the middle plate for the region of its movement \begin{align*}\triangle x \le x_o\end{align*}. Give your answer in terms of the constants given. (Hint: In this region both springs are providing opposite compression forces.)
 Find the force, \begin{align*}F\end{align*}, on the middle plate for the region of its movement \begin{align*}\triangle x \ge x_o\end{align*}. Give your answer in terms of the constants given. (Hint: In this region, only one spring is in contact with the middle plate.)
 Graph \begin{align*}F\end{align*} vs. \begin{align*}x\end{align*}. Label the values for force for the transition region in terms of the constants given.
 You design a mechanism for lifting boxes up an inclined plane by using a vertically hanging mass to pull them, as shown in the figure below. The pulley at the top of the incline is massless and frictionless. The larger mass, \begin{align*}M\end{align*}, is accelerating downward with a measured acceleration a. The smaller masses are \begin{align*}m_A\end{align*} and \begin{align*}m_B\end{align*} ; the angle of the incline is \begin{align*}\theta\end{align*}, and the coefficient of kinetic friction between each of the masses and the incline has been measured and determined to be \begin{align*}\mu_K\end{align*}.
 Draw free body diagrams for each of the three masses.
 Calculate the magnitude of the frictional force on each of the smaller masses in terms of the given quantities.
 Calculate the net force on the hanging mass in terms of the given quantities.
 Calculate the magnitudes of the two tension forces \begin{align*}T_A\end{align*} and \begin{align*}T_B\end{align*} in terms of the given quantities.
 Design and state a strategy for solving for how long it will take the larger mass to hit the ground, assuming at this moment it is at a height \begin{align*}h\end{align*} above the ground. Do not attempt to solve this: simply state the strategy for solving it.
 You build a device for lifting objects, as shown below. A rope is attached to the ceiling and two masses are allowed to hang from it. The end of the rope passes around a pulley (right) where you can pull it downward to lift the two objects upward. The angles of the ropes, measured with respect to the vertical, are shown. Assume the bodies are at rest initially.
 Suppose you are able to measure the masses \begin{align*}m_1\end{align*} and \begin{align*}m_2\end{align*} of the two hanging objects as well as the tension \begin{align*}T_C\end{align*}. Do you then have enough information to determine the other two tensions, \begin{align*}T_A\end{align*} and \begin{align*}T_B\end{align*}? Explain your reasoning.
 If you only knew the tensions \begin{align*}T_A\end{align*} and \begin{align*}T_C\end{align*}, would you have enough information to determine the masses \begin{align*}m_1\end{align*} and \begin{align*}m_2\end{align*}? If so, write \begin{align*}m_1\end{align*} and \begin{align*}m_2\end{align*} in terms of \begin{align*}T_A\end{align*} and \begin{align*}T_C\end{align*}. If not, what further information would you require?
 A stunt driver is approaching a cliff at very high speed. Sensors in his car have measured the acceleration and velocity of the car, as well as all forces acting on it, for various times. The driver’s motion can be broken down into the following steps: Step 1: The driver, beginning at rest, accelerates his car on a horizontal road for ten seconds. Sensors show that there is a force in the direction of motion of \begin{align*}6000\;\mathrm{N}\end{align*}, but additional forces acting in the opposite direction with magnitude \begin{align*}1000 \;\mathrm{N}\end{align*}. The mass of the car is \begin{align*}1250 \;\mathrm{kg}\end{align*}. Step 2: Approaching the cliff, the driver takes his foot off of the gas pedal (There is no further force in the direction of motion.) and brakes, increasing the force opposing motion from \begin{align*}1000 \;\mathrm{N}\end{align*} to \begin{align*}2500 \;\mathrm{N}\end{align*}. This continues for five seconds until he reaches the cliff. Step 3: The driver flies off the cliff, which is \begin{align*}44.1 \;\mathrm{m}\end{align*} high and begins projectile motion.
 Ignoring air resistance, how long is the stunt driver in the air?
 For Step 1:
 Draw a free body diagram, naming all the forces on the car.
 Calculate the magnitude of the net force.
 Find the change in velocity over the stated time period.
 Make a graph of velocity in the \begin{align*}x\end{align*}direction vs. time over the stated time period.
 Calculate the distance the driver covered in the stated time period. Do this by finding the area under the curve in your graph of (iv). Then, check your result by using the equations for kinematics.
 Repeat (b) for Step 2.
 Calculate the distance that the stunt driver should land from the bottom of the cliff.
Answers
 \begin{align*}3.6 \;\mathrm{kg}\end{align*}
 \begin{align*}\mathrm{g} \sin \theta\end{align*}
 b.\begin{align*} 20\;\mathrm{N}\end{align*} c. \begin{align*}4.9\;\mathrm{N}\end{align*} d. \begin{align*}1.63 \;\mathrm{kg}\end{align*} e. Eraser would slip down the wall
 a. \begin{align*}1450\;\mathrm{N}\end{align*} b. \begin{align*}5600\;\mathrm{N}\end{align*} c. \begin{align*}5700\;\mathrm{N}\end{align*} d. Friction between the tires and the ground e. Fuel, engine, or equal and opposite reaction
 b. \begin{align*}210\;\mathrm{N}\end{align*} c. no, the box is flat so the normal force doesn’t change d. \begin{align*}2.8 \;\mathrm{m/s}^2\end{align*} e.\begin{align*} 28 \;\mathrm{m/s}\end{align*} f. no g. \begin{align*}69\;\mathrm{N}\end{align*} h. \begin{align*}57\;\mathrm{N}\end{align*} i. \begin{align*}40\;\mathrm{N}\end{align*} j. \begin{align*}0.33\end{align*} k. \begin{align*}0.09\end{align*}
 a. zero b. \begin{align*}kx0\end{align*}
 b. \begin{align*}f_1= \mu_km_1\mathrm{g} \cos\theta; f_2 = \mu_km_2\mathrm{g} \cos\theta\end{align*} c. Ma d. \begin{align*}T_A= (m_1 + m_2) (a + \mu \cos\theta)\end{align*} and \begin{align*}T_B = m_2a + \mu m_2 \cos\theta\end{align*} e. Solve by using \begin{align*}d = 1/2at^2\end{align*} and substituting \begin{align*}h\end{align*} for \begin{align*}d\end{align*}
 a. Yes, because it is static and you know the angle and \begin{align*}m_1\end{align*} b. Yes, \begin{align*}T_A\end{align*} and the angle gives you \begin{align*}m_1\end{align*} and the angle and \begin{align*}T_C\end{align*} gives you \begin{align*}m_2, m_1 = T_A \cos 25/\mathrm{g}\end{align*} and \begin{align*}m_2 = T_C \cos 30/\mathrm{g}\end{align*}
 a. \begin{align*}3\end{align*} seconds d. \begin{align*}90 \;\mathrm{m}\end{align*}