Solving Force Problems in Physics using vectors The following are examples of how to solve a variety of problems using forces in the most straightforward manner possible. Each step in the problem solving process is clearly defined. The goal of this guide is to teach through practice the problem solving steps involved in these problems. Try to solve each problem on your own, reading only what to do on each step and not the solution.
Example 1 : Consider the problem below in which a block of mass m is sitting on a frictionless surface inclined at an angle of θ degrees. Determine the acceleration of the block. Solution : As always, the first thing to do is to draw the correct picture and DEFINE your coordinate system. It is quite common in the types of problems with inclined planes and pulleys to tilt the coordinate system. It makes no difference in the final answer which orientation you choose – as long as you are self consistent. The reason to tilt it is that it makes forces perpendicular or parallel to the plane (like normal force, tension, friction, etc) easy to split into x and y components. After your problem is well understood, the next step is two locate ALL of the forces acting on our block. In this case there are only 2 – the normal force and the force due to gravity (here denoted W indicating weight). You should then write out each of these forces in the most generic way possible (in terms of the magnitude of the force), like so : =〈0, N 〉 (Because the way our coordinate system is defined the y direction is normal to the N surface). =〈−W sin ,−W cos 〉 (Take note of the geometry of the similar triangles tells us that the two W angles marked θ are the same and that both of the x and y components are negative). The signs of the components are essential to getting the proper result.
We know that the net force on an object is simply the sum of all the forces on it, so we simply need to add the two vectors we have just discussed : NET = N W =〈−W sin , N −W cos〉 F
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NET =m a NET . One can see that IF the block is to Newton's Second Law tells us that F accelerate, it will only do so in the x direction (in this specific case the negative x direction). We can use this to write an equality : NET =〈−W sin , N −W cos 〉=m a NET =m〈a NET ,0 〉 F → 〈−W sin , N −W cos 〉=〈m a NET , 0〉 Finally, in order to make this equality work, the x components must be equal to one another, and the y components must be equal to one another. This gives us two equations : −W sin =m a NET and N −W cos =0 We know that the force due to gravity (near the surface of the earth) is given by W =mg . At this stage we can plug in this value to our formula : −W sin =−mg sin =m a NET → a NET =−g sin . Note that we can see that this is a negative quantity (pointing down the slope). We can then rewrite the acceleration in vector form : a NET =〈−g sin , 0〉 . Also from this problem we can see that the normal force does not affect the acceleration, but we can solve for N using the second equation that we made : N −W cos =0 N =W cos =mg cos
Example 2: For the second example problem we will consider the exact same situation as we did in example 1, but we will include the force due to friction. The coefficient of kinetic friction will be μk and the coefficient of static friction will be μs. Solution : The first thing to do is to draw the correct picture. As before, the normal force is always drawn orthogonal to the surface of the plane, and the weight vector points straight down (toward the center of the earth). The friction vector (F) will always be parallel to the surface (again, using the tilted coordinate system as described ensures that it will then only have an x component). You can draw the friction vector in either direction (down the slope or up it), if you get a negative value for the magnitude of F that simply tells you that you picked the direction opposite of the correct one. Remember that friction always OPPOSES motion (it will point in the opposite direction of the direction that the block will be moving). It is for that reason that I have Problem Solving Guide for Forces in Physics
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drawn it pointing up the slope (as it was proven in the previous problem that the block will accelerate in the negative x direction). After your problem is well understood, the next step is two locate ALL of the forces acting on our block. In this case there are three. You should then write out each of these forces in the most generic way possible (in terms of the magnitude of the force), like so : =〈0, N 〉 (Because the way our coordinate system is defined the y direction is normal to the N surface). =〈−W sin ,−W cos 〉 (Take note of the geometry of the similar triangles tells us that the two W angles marked θ are the same and that both of the x and y components are negative). The signs of the components are essential to getting the proper result. =〈 F , 0〉 (Because the friction must be parallel to the plane). F We know that the net force on an object is simply the sum of all the forces on it, so we simply need to add the two vectors we have just discussed : NET = N W F =〈−W sin F , N −W cos 〉 F NET =m a NET . One can see that IF the block is to Newton's Second Law tells us that F accelerate, it will only do so in the x direction (in this specific case the negative x direction). We can use this to write an equality : NET =〈−W sin F , N −W cos 〉=m F a NET =m〈a NET , 0〉 〈−W sin F , N −W cos 〉=〈 m a → NET , 0 〉 Finally, in order to make this equality work, the x components must be equal to one another, and the y components must be equal to one another. This gives us two equations : −W sin F =ma NET and N −W cos =0 We know that the weight of an object (near the surface of the earth) is always given by W =mg . Using the second equation above this gives us an expression for the normal force : N −W cos =0 N =W cos =mg cos . We also know that the magnitude of the force due to friction is always given by F = N . So solving for N in the previous part gives us something to substitute for F : F = N = m g cos . Now we will revisit the first equation (derived from the x components) : −W sin F =ma NET and substitute our relationships of W and F : −mg sin mg cos =m a NET It is useful to note that the masses cancel, leaving us with a final expression for the acceleration : a NET =−g sin g cos
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Example 3 : In this problem we will consider the case of a few blocks (of different masses) which are connected to one-another by ideal, massless strings being pulled as shown. Consider the floor to be frictionless. Find the tension in the rope connecting m2 and m3. Solution : As always, first be sure that you can draw the proper picture (it's an excellent measure of how well you understand what is being asked). In this problem we are lucky enough to have to only worry about the x direction (and so vectors will have only 1 number in them). One of the most useful pieces of information in this problem is that the the masses are connected by ideal strings (strings that don't stretch). This means that however quickly the first block gets pulled, the second and third blocks will be pulled along at the same speed. We can use this to say that the acceleration of each of the blocks will be the same and the system can be treated as though it were simply a large block being pulled by a force F (as shown in the figure below). You can see that if that were the case it would be very easy to find the acceleration of the system, simply using NET =m a NET Newton's Second Law : F F NET F = x (remember the hat m m1m 2m3 over the x is just to tell us that this acceleration is in the x direction). → a NET =
Now we know the rate at which each block will be accelerating, so we can easily find the net force on each individual block. Because we are looking for the tension in the string connecting m2 and m3 I have decided to look at block m3. It is important to note that you may choose m2 as well, the choice is completely yours. If we look only at the forces on m3, we can see that there are two : the tension T and the force with which we are pulling F. We then write out each of these vectors (again, these vectors are just in one direction, so it's pretty easy) : T =−T x =F x F We can find the net force on this block by simply adding up each of the forces : NET =T F = −T F x F NET =m a NET . Because we already know the From Newton's Second Law we know that F Problem Solving Guide for Forces in Physics
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F x m1m2 m 3 We can now solve for the tension in the string connecting blocks m2 and m3 : m3 T = F− F . m1m2m3 acceleration of block 3, we get :
NET = −T F x =m 3 F
[If you did the problem by considering block 2 you will have gotten something different than this, but m3 m1m2 =F don't worry, T = F 1− You can see these expressions are m1 m2m3 m1m2m3 equivalent.]
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