F1= 50.05g F2= 134.88g F=93.09g

[answerhappygod]

F1= 50.05g
F2= 134.88g
F=93.09g

F1 50 05g F2 134 88g F 93 09g 1 (42.28 KiB) Viewed 29 times

F1 50 05g F2 134 88g F 93 09g 2 (65.55 KiB) Viewed 29 times

F1 50 05g F2 134 88g F 93 09g 3 (52.44 KiB) Viewed 29 times

The purpose of this experiment is to practice the addition of vector graphically and analytically and to compare the results obtained by these two methods Apparatus: Cenco force table with pulleys, metal ring, string, mass hangers, masses, ruler, protractor The force table provides a means for applying known forces at one or more points and an various directions in the horizontal plane. The forces are the tensions in strings which pass over pulleys attached to the rim of the circular table and from which masses are hung. Terminology: Objects called "weights" have both a mass measured in kilograms for grams) and a weight measured in Newtons (or pounds). Weight is a measure of force (W = mg), NOT a measurement of mass. In the Sl (metric) system, weights are sometimes designated by their mass. For example, a 100 gram "weight" is referred to by its mass of 100 grams.
Theory: If several forces act on a body, the vector sum of these forces governs the motion of the al forces acting on it is zero (the vector sum is called the resultant force). The body is at rest or will move with a constant velocity (if originally in motion) if the vector sum of body. According to Newton's 1" w of Motion the body will remain at rest (if originally then said to be in translational equilibrium them so that the body is in translational equilibrium, and we determine how nearly these In this experiment, we consider forces acting on a small body (a metal ring), arranging will be measured in units called "gram-weight (mwt). One gmwt is the force of gravity forces satisfy the translational equilibrium condition that their vector sum is zero. Forces on one gram of mass. Therefore a mass of Mgrams will correspond to a force of Mgmwt. Ask your instructor to discuss this further if you are still confused Procedure: at not-too-symmetrical angles (otherwise the case becomes trivial). Placing one of the Note the strings already tied to the small ring at the center of the table. Clamp the pulleys pulleys at the O- mark of the force table may be helpful to you but it is not necessary. Add weights to the hanger at the end of each string and adjust the position of the pulleys it necessary until the ring comes to rest at the exact center of the table. Tap or jiggle the force table to reduce the effect of friction on the pulleys. Since the strings are radial from the center of the force table (especially if the pulleys are properly aligned) the direction of the forces may be read directly from the degree scale of the table. Record the angular position of the pulleys and the weight hanging from each string. (Remember to include the hanger weight in the overall weight.) Determination of the Resultant Force: Determine the vector sum of the horizontal forces acting on the ring in two ways: graphically, and analytically, i.e. by summing trigonometric components, (in the following discussion vectors are shown in boldface letters.) Graphical Method: Let Fi, F2, and Fs be the forces applied horizontally by the strings attached to the ring. Find the resultant Fr of these forces graphically, using a tail-to-head diagram (polygon of forces). Make your plot as large as possible, filling a page of your lab notebook. To accomplish this choose a suitable scale and be sure to indicate it on your drawing leg. 1.0 cm = 10 gmwt). Also indicate the reference direction you use (the direction from which you are measuring your angles). Be as precise as you can in using your ruler and protractor to draw lengths and angles.
Graphical Method continued): resultant vector due to friction in the pulleys or drawing errors. Determine the magnitude In "theory the magnitude Fall of Fa should be zero; however you will get a small Fal in gmwt and its direction in degrees. Analytical component) Method: Calculate (using trigonometry) the x- and y-components of each of the three forces based on your measured values of the magnitude and direction of each force. Start by assigning x- and y directions. It may be helpful if the direction of the x-axis is also the direction of one of the three forces. Calculate the x- and y. components of the resultant force (Fins and Fow) by calculating the algebraic sum of the three x-components and the three y-components, respectively Calculate the magnitude Fal and direction or of the resultant force from these components. Be sure to use the same reference axes in both methods Error Analysis of Results: The resultant force FR should be zero, of course, since the ring is not accelerating, However, your own result for Fx may be small but not zero. To get a sense of "how large" this non-zero experimental resultant is, compare its magnitude with the average of the magnitudes of the three horizontal forces in your set up. Do this by calculating the ratio of Fr to the average of the three magnitudes and state your answer as a percentage. Repeat for both graphically and analytically obtained results for FR. To assess the accuracy of your graphical Fr, compare it to your analytical Fr, by computing the percent difference of the two magnitudes. (Note: the denominator in this calculation will be the analytical value).