The Force Table. by Dr. James E. Parks

Transcription

1 b Dr. James E. Parks Department of Phsics stronom 401 ielsen Phsics Building The Universit of Tennessee Knoville, Tennessee Copright Ma, 000 b James Edgar Parks* *ll rights are reserved. o part of this publication ma be reproduced or transmitted in an form or b an means, electronic or mechanical, including photocop, recording, or an information storage or retrieval sstem, without permission in writing from the author. Objective: The objectives of this eperiment are: (1) to stud learn vector addition, () to learn how to resolve vectors into their components, (3) to learn how to find the magnitude direction of a vector from its components, (4) to learn how to find the balancing force for a bod that has two or more vector forces eerted on it. Theor Phsical quantities commonl are of two different mathematical tpes, scalars vectors. Scalar quantities are quantities in which onl the magnitude of the quantit is needed to epress its function. Vector quantities require both a magnitude direction to completel epress their characteristics. Common scalar quantities are time, mass, energ, temperature. Common vector quantities are displacement, velocit, acceleration, force, momentum. Scalar quantities can be added subtracted algebraicall using onl their values. However, vectors must be added vectoriall so that both their magnitude direction is taken into account. vector is specified b stating its magnitude direction relative to some coordinate sstem. In a Cartesian coordinate sstem, the direction can be specified b the angle the vector makes with respect to the ais. Vector quantities are distinguished from scalar quantities b placing an arrow above the smbol or b printing the smbol in bold tpe. Here we will use arrows, since it is much easier when using pen paper. For eample, two vectors ma be written as B. It is understood that these quantities have a direction associated with their magnitude. The smbols B without the arrow then means just the magnitute of the vectors B.

2 In order to add vectors in a practical manner, vectors can be resolved into two orthogonal components which when added together will be equal the vector. For eample, the vector can be resolved into a component along the positive ais one component along the positive ais. The direction of the positive ais can be designated b i the direction of the positive ais can be designated b j. These smbols are called unit vectors have unit value are used to specif the directions. The vector can then be resolved into its components written as = i+ j. The direction of can be specified b the angle it makes with respect to the positive ais. The components can be found from the trigonometric relationships are given b = cos (1) = sin () where represents just the magitude of the vector. Similarl for a vector B B=B i+b j (3) B =B cos (4) B B=B sin (5) Just as the vectors ma be resolved into their orthogonal components, the components ma be combined to reconstitute the vector as a magnitude direction. The magnitudes of vectors B, B, are given b B = + (6) B= B +B (7) their directions are specified b their angles with respect to the ais given b =rctan B B=rcTan B (8) (9)

3 Vectors B ma be added together to produce a sum vector, C,called the resultant, C=+B. (10) Just as B can be resolved into components, C can also be resolved into components, as a result, C=C i+c j (11) C = +B, (1) C = +B (13) C= +B i+ +B j. (14) ( ) ( ) Equations simpl state that the components of C are the algebraic sum of the components of B. s was the case for B, the magnitude direction of the vector C is C C are given b the equations ( ) ( ) C= C +C = +B + +B (15) C + B C=rcTan = rctan. (16) C + B For vectors i, ( i=1 to ) added together the resultant vector, generalizing Equations 15 16, so that R, can be found b R= R+ R= i+ i i= 1 i= 1 R =rctan (17) = rctan R i= 1 R i= 1 i i. (18) ewton s Second Law states that if a bod or mass is in equilibrium, then the sum of all the forces acting on the bod must be zero, or in equation form, 3

4 F=0 i (19) i=1 Since force is a vector quantit, this means that the sum must be the vector sum of the forces. When a bod is in equilibrium with several forces acting upon it, an one of the forces must be equal in magnitude opposite in direction to a resultant vector that is the sum of the remaining forces. n one force can balance out the remaining forces which can be summed as one resultant force, F R. This balancing force is called the equilibrant force, F E, F+F R E = 0. (0) so that F = F E In this eperiment 3 forces will be specified to be applied to a ring on a force table. These forces will be summed analticall to find their resultant force, F R. Then the equilibrant force, F E, will be found using Equation 1. The equilibrant force will be determined eperimentall b finding the force that is required to balance the ring when the 3 assigned forces are applied to it. This measured equilibrant force is then compared with the calculated equilibrant force. If F R is the resultant force from the sum of the three assigned forces, then R (1) F = F + F + F, () R 1 3 F = ( F + F + F ) iˆ+ ( F + F + F ) ˆj, (3) R ( ) ( ) F = F + F + F + F + F + F, (4) R FR F1 + F + F3 R =rctan = rctan. (5) FR F1 + F + F3 The equilibrant force,, F E, then will be given b F = ( F + F + F ) iˆ ( F + F + F ) ˆj E (6) ( ) ( ) F = F = F + F + F + F + F + F (7) E R

5 F + = F + F + F +. (8) R 1 3 E =rctan 180 rctan 180 FR F1 + F + F3 is the force to balance the given forces to compare with the measured force. pparatus The apparatus for the force table eperiment is shown in Figure 1. The apparatus consists of a force table, level, masses. The force table is circular with a graduated circular scale to convenientl determine the angles directions of the forces that are applied to a center ring. The center ring is positioned at a post at the center of the table to help determine when the forces on the ring are balanced to prevent the ring from moving off the table before it is balanced. Forces are applied to the center ring b attaching one end of a string to the ring passing the string over a pulle attaching masses to the other end. The table should be level so that there is no unbalanced gravitational force acting on the ring. Figure 1. Force Table with weight set. Procedure 1. Begin the eperiment b opening up an Ecel spreadsheet. Tpe the row column headings in column row 1 as shown in the eample shown in Table 1. Use the cop paste operations to enter repeated tet. The cells with #### are cells that 5

6 ou will make calculations or will enter measured values. The shaded cells will have no information should be left blank shaded. Table 1 B C D E 1 Forces Magnitud e ngle X Component Y Component Force # =B*COS((C/180)*PI() ) =B*SI(C/180*PI() ) 3 Force # #### #### 4 Force # #### #### 5 Sum of #### #### Components 6 Resultant Force R #### =T(E5/D5)/PI()* Equilibrant Force E #### #### 8 Measured Force 9 10 Force # #### #### 11 Force # #### #### 1 Force # #### #### 13 Sum of #### #### Components 14 Resultant Force R #### #### 15 Equilibrant Force E #### #### 16 Measured Force Force # #### #### 19 Force # #### #### 0 Force # #### #### 1 Sum of #### #### Components Resultant Force R #### #### 3 Equilibrant Force E #### #### 4 Measured Force 5 6 Force # #### #### 7 Force # #### #### 8 Force # #### #### 9 Sum of #### #### Components 30 Resultant Force R #### #### 31 Equilibrant Force E #### #### 3 Measured Force Four sets of vectors are given, with each set having 3 forces to appl to the center ring. With each force, its magnitude direction is given. Tpe these values shown in Table 1 into our spreadsheet. 6

7 3. Set up the force table with these 3 forces being applied to the ring. Once this is done ou should appl a fourth force, the measured equilibrant force, to balance the ring. The easiest wa to do this is to start with the 3 forces being applied to the ring then sliding the pulle clamp around the circumference of the table while using one h to pull on the string observing the position of the pulle where the ring will be centered. fter the direction of the equilibrant force is found, ou can then load the weight hanger with enough weights to completel balance the ring around the center post. Record this force, the magnitude direction, as the measured equilibrant force in cells B8 C8. 4. Using the relationships given in Equations 1, compute the components of force #1 from its magnitude direction record the results in cells D E. This can be done automaticall in the Ecel spreadsheet in the following wa: In cell D tpe =B* then click on Insert on the top menu bar. Choose function a Paste Function window should appear. Choose Math & Trig from the Function categor list COS from the Function name list. Then click on OK. n enter umber bo will appear with a button at the far right h end. Click on this button another input line bo will appear. Click on cell C C should appear in this bo. dd *PI()/180 to the line so that it reads C*PI()/180 then click on the button at the far right of the input bo. This will return to the Function palette. Click on OK. This procedure multiplies the magnitude of the first vector times the cosine of the angle to find the component. The angle recorded in degrees in cell C has to be changed to radians to be a valid argument for the Ecel cosine function that is the reason that the angle is multiplied b π then divided b 180. The component can be found recorded in cell E similarl b choosing the sine (SI) function. 5. Repeat procedure 4 for forces # #3. 6. In cells D5 E5, record the sums of the components the components. 7. Using Equation 4, record the magnitude of the resultant vector in cell B6. Do this b tping in a formula use functions from the insert menu where appropriate. 8. Use Equation 5 to find the angle in degrees of the resultant vector record this value in cell C6. gain tpe in a formula use functions from the insert menu where appropriate. Ecel finds the angle in radians using the arctan function, T() should be converted to degrees. 9. In cells B7 C7 record the values of the magnitude angle of the equilibrant force as determined from the resultant vector in step 8 above. Use Equations 7 8 to find these results. 7

8 10. Compare the magnitude of the measured equilibrant force with its calculated value. Compute the percent difference. 11. Compare the measured value of the angle of the equilibrant force with its calculated value. In this case, it doesn t make sense to compute a percent difference. If the angle of the calculated angle is more than 360, substract 360 from the angle to reduce it to a normal angular value. 1. Repeat these procedures for the remaining sets of forces. 8