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# LABORATORY FOR PHY 302K PHY 102M

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This 8 page Class Notes was uploaded by Malvina Orn on Monday September 7, 2015. The Class Notes belongs to PHY 102M at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 156 views. For similar materials see /class/181836/phy-102m-university-of-texas-at-austin in Physics 2 at University of Texas at Austin.

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Date Created: 09/07/15

Newton 5 1 and 7 Law and VectorAddition 21 2 NEWTON39S FIRST AND SECOND LAW AND VECTOR ADDITION The world evolves in a three dimensional space and therefore when we specify the velocity of an object we must give both its magnitude and its direction in order to adequately describe is motion Newton39s rst law says that an object maintains is state of uniform motion constant velocity if the net force the resultant force acting on it is zero If it is moving it continues moving in a straight line with constant speed If it is at rest it remains at rest If several forces act simultaneously on an object it will still be in equilibrium if the vector sum the resultant of the forces is zero If the resultant force on an object is not zero then Newton39s second law tells us that the object will be accelerated and its velocity will change Forces displacements velocities and accelerations are vector quantities They have directions as well as B magnitudes When two or more forces act on a single object the object responds as though a single resultant A B A force the vector sum of all the forces acting on the object has been exerted on it In this experiment we will R determine the forces necessary to place an object in equilibrium and we will verify that the net force the vector sum of all the forces is zero Fig 21 21 BACKGROUND DISCUSSION A vector can be represented graphically in the form of an arrow which has a length proportional to the magnitude of the vector quantity and a direction which corresponds to the direction of the vector quantity The vector arrow can be moved anywhere on the graph without changing its properties as one does not change its length or its direction Two vectors A and B are added by joining the tail of the B arrow to the head of the A arrow The resultant vector RAB is an arrow drawn from the tail of A at origin 0 to the head of B as illustrated in A Fig 21 The length ofR can be measured with a Ay A ruler and the direction of R can be measured with a protractor Instead of drawing a graph of the vectors and measuring the resultant the techniques of trigonometry can be used to calculate the length and direction Figure 22 shows how the vectors A Fig 2 2 Newton s 1 and Law and VecrarAddman 22 and B can each be decomposed into two vector components A Acos8A Ay Asin8A 21 and BK B 00585 By B sin85 22 where A and B are the lengths of A and B respectively and 8A and 85 are the angles between each vector and the positive x direction If we assume that A represenw a force of 100 N acting in a direction 531 above the positive x direction Eq 21 yields AX100N xcos531 600N and Ay100N xsin531 800 N Ifwe further assume that B represents a force of 1118 N acting 210 from the positive x direction Eq 22 yields BX118N xcos210 7986N and By118N xsin210 7559 N Since A and 3 both lie along the same direction as shown in Fig 23 they can be added algebraically to nd the x component of R R AX 3 600 N 7968 N 7368 N in the same way Ry Ay By 800 N 7559 N 241 N The graphical addition of these components is illustrated in Fig 23 The tail of BX is attached to the head of AK and R is drawn from the tail of AX to the head of BX Similarly the tail of 339 2 3 Byis attached to the head of Ay and Ry is drawn from the tail of Ay to the head of By The resultant vector Ris drawn from the tail of RX to the head of Ry It points le ward since R is negative and upward sinceRy is positive We can nd R the magnitude of the resultant vector R by the Pythagorean theorem RquotRRyz l68NZJr241NZ 440 N 23 The angle 8 between R and the xaxis can be found from trigonometry tan8RyRX 241N7368N70655 24 Newton s 1 and Law and VectorAddition 23 from which one obtains 8 33quot There is some ambiguity regarding the proper quadrant for the angle when using the arctan function so you have to look at Fig 23 to see that 33 is the angle between R and the negative x direction If there are several Vectors to be added such as R A B C D we can nd the componenw of each Vector using the method of Eq 21 We then sum them RXAXBXCXDX 25 RyAyByCyDy 26 and then apply Eqs23 and 24 as before 22 THE EXPERIMENT The apparatus is illustrated in Fig 2 4 It consists of a round force table with its circumference marked off in degrees Pulleys can be attached to the circumference at various angles by means of clamps so that weights can be used to apply forces through strings tied to a ring at the center of the table You will hang two or more weights from the ring and determine the sizes and angles which lead to equilibrium of the ring 23 PROCEDURE Hang two masses of approximately 200 gram total each at 0 and 90quot respectively Put a third pulley at 225 and see how much mass you have to suspend from it to get equilibrium Always make sure that the strings run in a straight line from the central hole to the pulley and that the pulley is pointed straight toward the central hole Measure using the triple beam balances and record the masses and angles in the excel spreadsheet N This time you will use a total of four different masses Hang m1 m 100 gat 0 mZ m 200 g at 80quot and m3 at 1N5quot where Nis the last digit of your student ID number and m3 is yet to be determined Hang a fourth mass m4 over a pulley whose angle you will vary Establish equilibrium by choosing different masses for m3 and mA and different angles for m When you establish equilibrium record the masses and angles in the excel spreadsheet Newton 5 1 and 7 Law and VecrarAddman U 2 4 Hang two masses from the force table mass mm at 0 and mass mm at 180 They should each have a total mass of N250 g does not have to be exact The ring is in equilibrium when it is centered over the hole in the middle of the force table Now change one of the masses by small amounts until the ring is out of equilibrium You will know it is out of equilibrium if when you give the ring a slight push towards the side with the heavier mass it starts to accelerate A er it is no longer in equilibrium record the values of mU and mm in the excel spreadsheet and print out all your data 90quot 180 0 00 270 Fig 25 24 INTERPREI ATION OF MEASUREMENTS 1 We will use the data from the 2string setup to get an idea of the tolerance of the force table Ideally equilibrium would only be achieved if the two masses in the 2string setup were exactly the same Since you found that the system was in equilibrium for a range of masses error sources such as friction in the pulleys or elasticity of the string must exert some force To get an approximation of how much force each string and pulley in the system can exert use the formula below to nd the mass tolerance of the 2string setup 5m imu 7mm 2 This value tells us over how large a range the masses can vary and still have the system in equilibrium We multiply by gravity to nd the range of force We have divided by two because we attribute half of this tolerance to each pulley This is obviously a rough approximation but the resulting formula for the force tolerance per pulley is Newton 5 I and 2 Law and VectorAddition N 5Fg5m Find the uncertainty in the angle 59 This will be half the smallest measurable angle For both the 3string and the 4string setups complete the following steps a Using the mass values you measured ll in the tables with the corresponding force on the ring Remember the force F on the ring is the tension T in string which must be equal to the weight of the attached mass For symbolically inclined we have FTmg the the b On a separate sheet of graph paper draw the freebody diagram of the ring as in Fig 25 Choose the XaXis of the drawing to correspond with 0 on force table and draw the lengths of the vectors to scale Remember you drawing the forces applied to the ring Your scale should be in Newton the are c Using Fig 21 as a guide add the vectors graphically on the piece of graph paper To do this you must draw the vector to scale use a different scale than what you used for the previous part 2b so you can fill the entire page with your vector addition joined headtotail and measure the length and angle of the resultant vector The units should still be Newton 1N 1 ms d Add the vectors algebraically The chart has spaces for you to break them i components and sum those components Then use the sums of components to find the magnitude and direction of the resultant e Find the relative percent error for the resultant obtained graphically and resultant obtained algebraically The relative percent error is defined as magnitude of the resultant divided by the average magnitude of component forces For example gtlt100 FlFzFs3 quot1 the set up Kg nto the the the the Using the force tolerance per pulley find the hypothetical force tolerance of Newton 5 I and 2 Law and VectorAddition Newton 5 I and 2 Law and VectorAddition 2 7 PHY 102M Lab Report Experiment 2 Newton s 1st and 2 d Law and Vector Addition NAME CLASS TIM l RAW DATA 121 Force 10 2 string setup Mass at 0 degrees m0 Mass at 180 degrees m180 2 EXPERIMENTAL RESULTS 221 2 string setup Mass uncertainty for each hanging mass 5m lm0miaol 2 Force uncertainty per pulley 5F g 5m Uncertainty of the angle 59 2b 3 string setup Attach freebody diagram of forces on the ring and a graphical vector addition of those forces to lab report on separate page Magnitude of graphical resultant R g Direction of graphical resultant 6g R Graphical relative percentage error g X 100 FlFzFg3 Newton 5 I and 2 Law and VectorAddition 2 8 PHY 102M Lab Report Experiment 2 Newton s 1st and 2 d Law and Vector Addition Magnitude of algebraic resultant Ra Direction of algebraic resultant 6n R Algebraic relative percentage error X 100 Fl F2 F 3 Measurement uncertainty of the resultant 35F 20 4 string setup Attach freebody diagram of forces on the ring and a graphical vector addition of these forces to the lab report on separate page Magnitude of graphical resultant Rg Direction of graphical resultant 6g R Graphical relative l error g X 100 E F2 Fg F44 Magnitude of algebraic resultant Ra Direction of algebraic resultant 6n error R X 100 Fl F2 Fs F44 Measurement uncertainty of the resultant 45F A Algebraic relative l 3 CONCLUSION SECTION This section contains answers to questions on this lab provided by your lab instructor

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