1 When we have forces acting in two dimensions, we need to understand how to find the mathematical answers when we add vector quantities that are not operating in one dimension? As a simple example, let’s imagine that we have a 1.5 kg book sitting on a table. One person will push on the book in the positive x direction with a force of 5.0 N while a second person pushes on the book in the positive y direction with a force of 3.0 N. Assuming that the table is frictionless (makes the problem simpler), what would happen to the book? +y +x

2 +y We need to “add” the vector quantities together to find the resultant vector. +x Graphically, we do this by drawing the vectors from tip to tail and then drawing a new vector that connects the tip of the first vector to the tail of the second vector. +x +y Now we can think of this single vector as being the only force applied to the book. Therefore the book will move in the direction indicated by the vector and will accelerate according to the equation F = ma

3 How can we solve this problem mathematically instead of graphically?We need to use relationships that we first learned about in geometry. a Notice that in this case, the two forces were acting 90o relative to each other. This means that our diagram represents a right triangle. Therefore we can use the Pythagorean Theorem (c2 = a2 + b2) to find the length of the resultant vector. From our initial example: a is 5.0 N, b is 3.0 N and c will be the resultant. c2 = a2 + b2 = = 34 and c = 5.8 N Try practice problems 1 and 2 on page 121.

4 What happens when the forces (or displacement) is not at a 90o angle?b c  We still need to use relationships that we first learned about in geometry. If one or more angles in the diagram is known, then we can use the Law of Cosines or the Law of Sines. Law of Cosines: c2 = a2 + b2 – (2ab)cos () Law of Sines: 𝑐 sin⁡(𝜃) = 𝑎 sin⁡(𝛼) = 𝑏 sin⁡(𝛽) As an example: a is 5.0 N, b is 3.0 N,  = 135o and c will be the resultant. c2 = a2 + b2 – (2ab)cos() = – (2*5.0*3.0)cos(135) = 55.2 and c = 7.4 N note that using the Law of Sines now gives us b. Try practice problems 3 and 4 on page 121.

5 More Vector Issues Component Vectors: all two dimensional vectors can be broken down into component vectors which are parallel to the x-axis and parallel to the y-axis. +y A Ay +x Ax Note that forming the component vectors from a given vector is just the opposite of the process of finding the resultant vector from two vectors. We choose to use two vectors that are 90o relative to each other so that they can then represent the x-axis and the y-axis components respectively. By convention, the angle of a vector relative to the x-axis will be used to designate the direction of the vector. The angle is measured counterclockwise from the x-axis.

6 Angle Conventions +y +y A A Ay Ay +x +x Ax Ax Angle will be between 0o and 90o Angle will be between 90o and 180o +y +y Ax Ax +x +x Ay A Ay A Angle will be between 180o and 270o Angle will be between 270o and 360o

7 Formulas involving the vector’s angle and the component vectors.Ax = A cos () Ay = A sin ()  = tan-1( 𝐴𝑦 𝐴𝑥 ) +y A Ay  +x Ax Example: A vector has a magnitude of 12 and a direction of 75o (i.e. ( = 75o)). Find the component vectors. Example: A vector has Ax = – 7.3 and Ay = Find the magnitude and direction of the resultant vector.

8 Adding Multiple VectorsExample: You travel 3.5 km at 20o, then you travel 1.8 km at 60o, and then you travel 4.5 km at 10o. Find your location (distance from origin and angle from x-axis). Click to see solution +x +y 1 = 20o 2 = 60o 3 = 10o Ax = A cos () Ay = A sin ()  = tan-1( 𝐴𝑦 𝐴𝑥 ) Ax1 = 3.5 cos(20) = Ax2 = 1.8 cos(60) = Ax3 = 4.5 cos(10) = 4.43 Rx = Ax1 + Ax2 + Ax3 = = 8.62 Ay1 = 3.5 sin(20) = Ay2 = 1.8 sin(60) = Ay3 = 4.5 sin(10) = 0.78 Ry = Ay1 + Ay2 + Ay3 = = 2.81 R2 = Rx2 + Ry2 = = R = 9.06 = 9.1 km  = tan-1(2.81/8.62) = 18o

9 Practice Problems 5-10 on page 125

10 Friction: Friction is a force that resists motion.There are types of friction: Kinetic and Static Kinetic Friction occurs when two object are in contact with each other and one object is moving relative to the other. Imagine pulling on a rope tied to the wall. If you pull hard enough, the rope will slip through your hands and you will feel the friction heating your hands up. Can you think of other examples? Static Friction occurs when two object are in contact with each other and the objects are not moving relative to the other. Imagine pushing a heavy rock. Even though you pushed hard, the rock did not slide across the ground. Why not? What force is keeping it from moving? If the rock was smooth and was sitting on ice, would it be easier to get it to move? Why or Why Not?

11 The amount of friction between two objects depends upon the type of surface the objects have (rough, smooth, etc… ), whether the objects are wet or dry (many liquids act as lubricants between two solid objects-like water on a road will make it easier for your tires to slide), what the objects are made of (cement versus glass, etc…), and the normal force between the objects (essentially how hard the objects are being pushed together). The coefficient of kinetic friction (mk) is how we can take the surface features of objects into account. By measuring the amount of force required to pull an object across a surface at constant velocity and then by increasing the weight of the object being pulled you can create a graph of the Ff, kinetic versus FN. The slope of this line is called the coefficient of kinetic friction. Kinetic Friction Force (Ff, kinetic) is equal to the coefficient of kinetic friction multiplied by the normal force (FN). In equation form: Ff, kinetic = mkFN

12 When two objects are not moving relative to each other, then the static friction force will be involved in the problem. Remember that if nothing is moving then we are at equilibrium. In the case were we are pushing on an object, but the object is not yet moving, then the force of friction is as large as it needs to be to keep the object from moving. If we push harder (still not moving), the static friction force will increase. Eventually if we push hard enough, the object will start to move. At the point where we just get the object to start to move, we can find the maximum static friction force. In equation form: Ff,static ≤ msFN The coefficient of static friction (ms) is found in a similar manner as the coefficient of kinetic friction. The less than or equal to sign (≤) in the equation means that the maximum static friction force will be achieved when the object just starts to move. Up until that point, the static force of friction will be less than the maximum-it will be as large as is required to keep the object from moving. Table 5-1 on page 129 has some coefficients of static and kinetic friction for some common materials. You DO NOT have to memorize these! Note that the coefficients of kinetic friction are all less than the coefficients of static friction for a given set of materials.

13 Example: You are pushing a 37Example: You are pushing a 37.0 kg wooden box across a wooden floor at a constant speed of 1.35 m/s. What force do you have to exert on the box to keep it moving at this speed (i.e. what is the pushing force Fp)? Note: since the box is moving at constant speed we know that we will be using the coefficient of kinetic friction and we know that we are at equilibrium (i.e. the net force on the box is zero). +x +y FN Fg Fp Ff, kinetic Fnet = 0 Since Fnet = 0, we know that FN = - Fg and that Fp = - Ff, kinetic = mkFN We can use this information along with the information given in the problem to find our answer. FN = - Fg = - (37.0 kg)( m/s2) = N Fp = mkFN = (0.20)(362.6 N) = N = 73 N

14 We should have some questions based on the previous example.Here are a few: Is the force we used to push the box at a constant speed of 1.35 m/s the same force we needed to get the box to start to move? If not, how hard did we have to push to get the box moving to begin with? If we wanted to move the box at a constant speed of 4.50 m/s would we have to push harder than we do to make it move at a constant speed of 1.35 m/s? What is the difference in the problem if we want to accelerate the box compared to just moving it at a constant speed? Your ability to understand and figure out the above questions and their implications is a good measure of how well you are developing your problem solving skills with respect to the Physics class. See the practice problems on page 128.

15 Example: What will happen when you push a 37Example: What will happen when you push a 37.0 kg wooden box across a wooden floor with a force of N? Note: Since this problem has the same set up as the previous example, we know that the force of friction (Ff, kinetic) for the box is - mkFN, and we know that Ff, kinetic = - 73 N. Therefore: Fnet = Fp + Ff, kinetic = 150.0N – 73 N = 77 N. +x +y FN Fg Fp Ff, kinetic Fnet = 77 N Since Fnet ≠ 0, we know that the box is accelerating! Remember that Fnet = ma Fnet = ma and a = 𝐅𝑛𝑒𝑡 m = 77 N 37.0 kg = 2.08 m/s2 = 2.1 m/s2 This is an example where the pushing force is not balanced by the friction force. See example problems on page 130.

16 Force and Motion in Two Dimensions: Imagine a situation where three forces are acting on an object as shown in the diagram below. FA FB FC +x +y We need to add the three vectors together to see what will happen to the object. Graphically this could be done by recognizing that force A and force B are acting parallel to the y and x axis respectively. Seeing this, we can break force C down into its components and then use them to complete the math. FB Looking carefully we see that FCx = - FA and that FB = - FCy Therefore: Fnet = 0 The object will not change its motion! FA FCx FC +x +y FCy

17 Another way to see that the net force for these three forces is zero is to move them until they are head to tail. If they make a closed triangle, then the net force is zero. FA FB FC +x +y +x +y FA FB FC Of course, we will also need to be able to do this type of problem purely mathematically if lengths and angles are given for each force.

18 A force that puts an object into equilibrium (makes the net force equal to zero) is called the equilibrant. +x +y FA FB Fnet Fequilibrant If forces A and B are acting on an object as in the diagram above, the net force is not zero and the object is not in equilibrium. If we apply a force that is exactly equal in size to the net force but is acting in the opposite direction (180o to it), then this new force would bring the object into equilibrium and the new force would be called the equilibrant. See the challenge problem on page 132.

19 What would happen if we set a box on a slopeWhat would happen if we set a box on a slope? How do we evaluate the problem? Ff FN Fg +x +y We need to know the angle of the slope. We need to know the mass of the object. We need to know the force of kinetic friction (assuming the box is moving). We need to know the force of static friction (assuming the box is at rest). Assume the angle is 140o. Assume the box weighs N. Assume the box and the slope are made of wood. Will the box slide down the slope?

20 This can be a very complex problem, but by choosing the direction of the axes correctly, we can simplify the math. If we choose the + x direction to be pointing down the slope, then the normal force will be in the + y direction and the problem will be simplified. Using this orientation: the normal force is in the + y direction, the friction force is in the – x direction, and if we break the weight into its component forces, we can find the net force acting on the box. Ff FN Fg +x +y Fnet = FN + Ff, kinetic + Fg = FN + Ff, kinetic + Fgx + Fgy However, FN = - Fgy so FN and Fgy will cancel out and the equation will become: Fnet = Ff, kinetic + Fgx

21 Fg FN Ff +x +y Fgx Fgy  From Geometry:  = 50o in this problem Using Law of Sines: 𝐅gx sin⁡(40) = 𝐅gy sin⁡(50) = 𝐅g 𝑠𝑖𝑛⁡(90) So 𝐅gx=𝐅g sin 40 = N =321.4 N and 𝐅gy=−𝐅g sin 50 =− N =−383.0 N - 𝐅gy=𝐅N so 𝐅N=383.0 N Ff, kinetic = - mkFN = - (0.20)(383.0 N) = N Fnet = Fgx + Ff, kinetic = N N = N Ff, kinetic = - mkFN = - (0.20)(383.0 N) = N Ff, static = - mstaticFN = - (0.50)(383.0 N) = N Since the net force is greater than the static friction force, the box will slide down the slope.

22 Fg FN Ff +x +y Fgx Fgy  Fnet = N Extension problem: based on the answer to the previous problem, how fast would the box be sliding down the slope after 5.0 s (assuming that is started from a stop at zero seconds). Fnet = N Fg = mg so m = Fg/g = N/9.80 m/s2 = 51.0 kg F = ma and a = F/m = N/51.0 kg = 4.80 m/s2 f = i + at = (4.80 m/s2) * (5.0 s) = 24 m/s down the slope See practice problems on page 135.