28 28 3 18 18 1 1 Force Table Clamp with Pulley Ring 21 19 18 17 15 23 PASCO scientific 1 13 2 25 1 11 2 1 27 8 9 28 8 29 3 31 3 33 3 3 3 35 Model ME-9447 FORCE TABLE 1 3 5 7 Ring Force Table Center Post Figure 2 Ring Method of ing Force Table Ring Method See Figure 2. To use this method, screw the center post NOTE: A string can be attached to the The up force until table it stops isso a that device it sticks with up which above the youtable. can analyze apasco number mass ofhanger forceby vectors wrapping at the string same time. It consists Place of the anring elevated over the circular post and tie platform one 3 cm graduated long in degrees several times to which (4 or 5) pulleys around can the notch be attached. the top Mass hangers string and to the slotted ring for masses each pulley. attached The strings to pieces must beof stringof run each over mass the hanger. pulleys and connected together provide long the enough necessary to reach over forces the (the pulleys. weight Place at eachthe end of the strings). string over a pulley and tie a mass hanger to it. Force Table Clamp with Pulley 21 19 18 17 15 23 PASCO scientific 1 13 2 25 1 11 2 1 27 28 29 3 3 3 3 8 7 9 8 Center Post 31 3 Model ME-9447 FORCE TABLE 5 33 3 35 1 3 Anchor Figure 3 Anchor Method of ing Force Table The defining characteristic of the table is that when the knot holding the strings is centered on the Anchor Method circular platform the forces are in equilibrium ( F = R = ). Associated with equilibrium is a vector known See as Figure the 3. equilibrant. Cut two cm lengths It is aof vector string and that, tie whenone added of the tolegs. another Put each vector of the other or series strings ofover vectors a will bring them these together vector(s) at their into centers equilibrium; (to form an "X"). i.e., Three R pulley and tie a mass hanger on the end of each string. x =, R y =, and R =. Obviously for a single vector of the ends will reach from the center of the table over it is a vector of equal magnitude in the oppsite direction pulley; the fourth will be threaded down through the (18 NOTE: away). A string For can a be system attached ofto multiple the vectors it is hole equal in the in magnitude center post to and act as opposite the anchor instring. direction to the PASCO resultant mass of hanger the by system wrapping of vectors. the string Screw the center post down so it is flush with the top several times (4 or 5) around the notch at the top When analyzing forces - or vectors in general - remember that vectors do not add like scalars. surface of the table. Thread the anchor string down of each mass hanger. through the hole in the center post and tie that end to Example 3 Given two force vectors v 1 = 2.8N at 3. from the positive x-axis (3. first quadrant) and v 2 = 4.N at 11. from the positive x-axis (7. second quadrant). Find the sum of (resultant) these vectors. Solution First, draw each vector from the origin in an x y coordinate system; do not change the direction of the vectors. They are shown below, along with their respective x and y components. The angles in parentheses above are the reference angles for each vector. Reference angles are the angles to the closest x-axis (positive or negative x-axis), so they are always and < 9. 1
The mangitude of the vector components comes from the relations v 1x = v 1 cos θ 1 = 2.8N cos 3. = 2.42N v 1y = v 1 sin θ 1 = 2.8N sin 3. = 1.N v 2x = v 2 cos θ 2 = 4.N cos 7. = 1.37N v 2y = v 2 sin θ 2 = 4.N sin 7. = 3.76N Note the negative sign on v 2x since v 2 is in the second quadrant. The components of the resultant are The magnitude of the resultant vector R is R x = v 1x + v 2x = 2.42N 1.37N = 1.5N R y = v 1y + v 2y = 1.N + 3.76N = 5.16N R = R x 2 + R y 2 = 1.5 2 + 5.16 2 = 5.3N and its direction θ = tan 1 R y = 5.16 tan 1 1.5 = 78 R x It should also be noted that this equation will always return a reference angle, and dropping the absolute value from the ratio will not give you an angle from to 3! It is up to you to determine the direction of the resultant based on the signs of its components. For example, a negative x component and negative y component would mean that the resultant lies in the 3rd quadrant. This method can be extended to any number of vectors. Apparatus Force table, Pulleys,, Slotted mass set. 2
Procedure Setup Cut off two cm pieces of string and tie them in the middle, forming an X. Tie the second to the first at the middle, then the second to the first at the middle; the knot should then stay in place. The top of the center post should be flush with that top of the force table; if not, adjust this first. Thread one of the strings through the post and tie it to one of the legs as shown in the figure above. This will keep the masses from falling off the force table if the system moves too far out of equilibrium. Note also in the figure that all strings are parallel to the top of the force table and go over the pulleys in straight ines at the same height. Be sure to check this when you set up your systems. The mass hangers are 5g, so remember to add this to any slotted masses you have on the hanger when you calculate the force (weight). The knot will not be perfectly centered every time you achieve equilibrium. It should move toward the center as you get closer but might end up stopping a few mm away. Is this equilibrium? You can move the knot away and see if it returns, but remember that F =. Try moving the knot so it is centered and let go; does it stay centered? If it does not try adding the smallest mass to the lighter side. If this pulls the knot away from the center then you had equilibrium. Data Collection 1. Test whether or not the components of a vector are equivalent to the vector itself. 2. Set up a system of two forces - F 1 from a hanger with 15g of slotted masses on it at., and F 2 its equilibrant. Make any necessary adjustments to the magnitude and direction of F 2 so that the system is in equilibrium. Record your results on the Data Sheet. 3. Calculate the components of F 1. Replace F 1 with these components - you will need a third hanger. 4. Test whether or not the equilibriant of a system of vectors is equivalent to the resultant of these vectors. 5. Set up a system of three forces; F 1 from a hanger with 1g at 155., F 2 from a hanger with 15g at 255., and F 3 the equilibrant of F 1 and F 2. Make any necessary adjustments to the magnitude and direction of F 3 so that the system is is equilibrium. 6. Calculate the magnitude and direction of the resultant of F1 and F 2. Compare this to the equilibriant. 3
Data Sheet F 1 1.52. F 2 F 1x (N) F 1y (N) F 1 1.3 155. F 2 1.52 255. F 3 R x R y R % difference in magnitudes of R and F 3 Angle between R and F 3 ( ) 4
Analysis 1. Did you maintain equilibrium between the components of F 1 and F 2? You should have, and it is for this reason that vector addition works using the components of the vectors. What do think is the largest source of error in the experiment? 2. What is the percent difference in the magnitudes of R and F 3? What should have been the angle between these vectors? 3. Can a vector have a negative magnitude? Explain. 5
Pre-Lab: Force Table Name Section Answer the questions at the bottom of this sheet, below the line (only) - continue on the back if you need more room. Any calculations should be shown in full. 1. Read the lab thoroughly; check the lab manual for any additional information. 2. What are the forces in this experiment? 3. What is an equilibrant? 4. What is a reference angle? 5. What is the reference angle for a vector whose direction is 23 from the positive x-axis? 6. What is the magnitude and direction of the resultant of forces F 1 = 7.N at. and F 2 = 4.6N at 17.? Show all work. 6