Name: Partner s Name: EXPERIMENT 1500-7 2D FORCE VECTORS INTRODUCTION A vector is represented by an arrow: it has a direction and a magnitude (or length). Vectors can be moved around the page without changing their meaning. For example, the two vectors on the right are equivalent. Addition of vectors can be accomplished in a number of different ways. We can add them graphically by putting the vectors head to tail, as depicted on the right. The easiest way of adding vectors algebraically is to decompose them into their components, and then add those components. It should be clear that we cannot simply add the lengths of vectors, since they will in general not be pointing in the same direction. The two ways of describing a vector (components or length and direction) are equivalent and are related by a set of formulae. However, if we want to add vectors algebraically, we need have to have the vectors components. In other instances, though, a description in terms of length and direction might be more useful. We will be using vectors in two dimensions, where a vector V is described by its length V V and its angle α, or by its two components, i.e. the projections of the vector onto the two axes, from length and direction to components use V V x y V cos V sin V, V x y. To get, to get from components to length and direction use 2 V Vx V Vy tan Vx 2 y. 1
Due to the ambiguities associated with the square root and the inverse tangent, be sure to always check that the length of a vector is positive (the components can be negative), and that the angle α is representing the correct quadrant. For example, when both components are negative, the sign will cancel and the calculator gives you an angle in the first quadrant, while the vector actually belongs to the third quadrant. All of this is best illustrated by working through the following example: A treasure chest can be found by starting at the big oak tree and walking 120 feet north-by-northwest, 50 feet northeast, then 25 feet south. What do you need to be doing to get to the treasure chest directly, i.e. in what direction and how far you would need to walk from the oak to the treasure chest? First, let s answer this graphically. Draw the three vectors on the grid with the help of pencil and protractor. As a scale, use 10 ft per box and start at in the middle of the lowest line. Take the angle to be in degrees North of East, and East to be the x-axis. First, translate directions into degrees: 1) 120 feet NNW = 120 ft at degrees 2) 50 feet NE = 50 ft at degrees 3) 25 feet S = 25 ft at degrees Now draw the vectors. Be sure to start the second vector at the tip of the first one, etc. Draw the resulting vector, i.e. the vector that leads directly to the treasure chest, and list its magnitude and direction below (use your ruler and protractor to measure the values of the length and angle). R = Angle of R = Now, let s answer the same question by using components. Use the formulae on the first page to decompose the vectors into their x (East) and y (North) components. Remember that components can be negative (show your work on the next page) 2
1) 120 ft NNW = ft East and ft North 2) 50 ft NE = ft East and ft North 3) 25 ft S = ft East and ft North Sketch each trip East and North as a set of little arrows in the grid below, in one continuous path. What is the net displacement East that you travel? ft What is the net displacement North that you travel? ft What is the net displacement if you were to do this in one straight leg? (show work below) ft In what direction would you need to travel to get there in one straight leg? (show work below) degrees. 3
Does this result agree with the previous result? If not, explain why. Break down the vector R from page 2 into its components. (show work below) R x = ft R y = ft Is this the same as the net east and north displacements? If not, explain why. Check with your instructor that you have the correct answers before continuing. THE FORCE TABLE A force is a push or a pull. In this lab, forces are created by hanging weights from strings. The strings go over pulleys and all attach to a central ring. This ring is pulled by the strings, with each string creating a different force on the ring. A force is most conveniently represented as a vector, as it has a direction and a strength or magnitude. We denote it as F. Since we have three forces caused by three weights A, B, and C, we can distinguish the force vectors created by each weight using subscripts: convenience, we will call F A simply A, we will call F B B, etc. F A F B, F,. For The direction of each force vector is simply the direction the string is pulling. The magnitude of the each force vector represents how hard the mass is pulling the string, which is proportional to the weight on the string. For a single weight on a string, the magnitude of a force vector is C F F m g mg. 4
where g = 9.8 m/s 2. Thus, a hanging mass of 200 g = 0.2 kg provides of force of 0.2kg*9.8 m/s 2 = 1.96 N. (The Newton is the SI unit of force, N=kg m/s 2.) The sum of several forces acting on a system is called the resultant force. Resultant just means vector sum: the total force you get for adding all the vector forces together. 1. Record the number on the card at your table and calculate the forces A, B and C from the masses (show your work). Table 1: Card: Label Direction Magnitude A B C 2. Using the strings and weight hangers, set up forces A and B on the force table at the proper angles and magnitudes. 3. Use a third string and weight to find the force that exactly balances the other two. You ll need to experiment with both the size of the weight and the direction of the string. All the strings should be in straight lines pointed away from the center pin on the force table when balance is achieved. In the table below, record the mass you needed to balance the other two, and the direction that you needed to hang the string to make it balance. 4. Once you have balanced the forces, estimate how well you had to the position the string in degrees to get the right answer. Record this as the uncertainty in direction. Then, add a few grams to the balanced weight until the forces are noticeably out of balance. Record this added force as the uncertainty on your magnitude. This is a measure of the experimental precision of your mass measurement. Repeat the above procedure for B C and A B C. 5
Table 2: Trial Mass Need ed (g) Balancing Force Direction Magnitude Uncertainty on Direction Uncertainty on Magnitude A B B C A B C CALCULATIONS On the next two pages, calculate the resultant force for the four vector sums, A B, B C and A B C, using the component method we practiced in the introduction. Beneath each graph, show all steps of the calculations in an orderly way that can be easily followed by another student. Sketch an approximate vector diagram on the graphs to illustrate the components and resultants for each problem. Obtain all angles as positive angles between 0 and 360 degrees. (Calculated results should be to 3 significant figures since your data are to 3 significant figures.) Summarize your calculations in the Table 3. Note that the resultant forces should be exactly the same size in the opposite direction as the ones you found experimentally before. Find the balancing force by taking the resultant force, but 180 degrees in the opposite direction. 6
A B : B C : 7
A B C : 8
Table 3: Trial Calculated Resultant Force Calculated Balancing Force A B Direction Magnitude Direction Magnitude B C A B C ANALYSIS Now we will compare the results of your calculations with your experiment. First, find how much your experimental result differs from your calculation by subtracting the calculated result from the measured magnitude. Second, copy the experimental uncertainty from Table 2. Record the values in the last two columns of Table 4. Next, do the same thing with the angles of the direction measurements. Table 4 Trial Direction Magnitude Disagreement with calculatio n Experimental uncertaint y Disagreement with calculation Experimental uncertainty A B B C A B C 9
Compare the disagreements with the experimental uncertainties. Make sure to comment on both magnitude and direction. Is your disagreement less than your experimental error (i.e. did your measurements agree with your calculations to within your error)? If not, why? What do you estimate the precision of the equipment to be and is your experimental error within the expected precision of the equipment? What does this imply about your experimental techniques? Are they more precise or less precise than the resolution of the instruments? 10