Lab 103: Translational Static Equilibrium-Force Table Objectives 1. To confirm t
ID: 3280548 • Letter: L
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Lab 103: Translational Static Equilibrium-Force Table Objectives 1. To confirm that the condition for static translational equilibrium is that the vector sum of forces is zero; 2. To experimentally test the vector nature of force; 3 To practice manipulating the vectors and attain better understanding of vectors: 4. To find unknown tensions and directions in a system of strings Introduction and Background connected to a central ring As stated moving with a constant speed in a straight line depending on its initial condition. by Newton's Laws of motion, a particle that experiences zero net force will either remain at rest or We know that physical quantities are generally classified as being either scalar or vector quantities. Force s e vector quantity, and it has both magnitude and direction. So when handling the force and trying to get th force, the manipulation rules of vectors have to be followed. please refer to the math book. I. Graphical Method-Triangle (Head to Tail) Method are generally two methods for the addition of force vectors. Following is a brief description. For details, Vectors are represented graphically by arrows. The length of a vector arrow (drawn to scale on graph paper) is proportional to the magnitude of the vector, and the arrow points in the direction of the vector. Shown in Figure are the additions of vector A and B, vector A, B, and C with vector Ri and R being the resultant vector 1 respectively Il. Analytical Method-Component Method Any vector could be decomposed into x and y components. The vector sum of any number of vectors can be obtained by adding the x and y components of the vectors. The magnitude of the resultant vector is then given by Re (R. R where R A+B,+C.. and R-A,+B, C.., and the direction of the resultant vector Bstan R/R.).Shown in Figure 2 is the case of two vectors. L aws of motion, if a body is acted upon only by concurrent forces (i.e., forces whose lines of at a point), the condition for static equilibrium is that the vector sum of the concurrent forces Then it follows that: I. When using the graphical method, one force vector has the same the sum of all the other force vectors and opposite direction to the sum of the vectors (see Figure Based on Newton's of must be zero. magnitude as HYS 111AExplanation / Answer
Case 1 : Ta = 10.29 N
theta = 5deg
theab = 175
thetac = 270 deg
then in vector notation
Ta = 10.29(cos(5)i + sin(5)j) [ where i and j are unit vectors along x and y directions]
Tb = |Tb|(cos(175)i + sin(175)j)
Tc = |Tc|(cos(270)i + sin(270)j)
so for static equilibrium
Ta + Tb + Tc = 0
hence
-0.99619|Tb| = -10.250
Theoretical values
Tb = 10.29 N
and
10.29sin(175) + 10.29sin(5) - |Tc| = 0
Tc = 1.7936 N
Measured values
Tb = 10.50 g
Tc = 1.8 g
% error
error in Ta = 0%
error in Tb = 2.04%
error in Tc = 0.35%
Case 2:
Ta = 1.3818
theta = 17
Tb = 2.7636
thetab = 142 deg
so from static equilibrium condition
1.3818(cos(17)i + sin(17)j) + 2.7636(cos(142)i + sin(142)j) + Tc(cos(thetac)i + sin(tehtac)j) = 0
hence
Tcsin(thetac) = -2.10544
Tccos(thetac) = 0.8563
hence
Theoretical values
thetac = -67.867 deg + 360 = 292.133 deg
Tc = 2.272
Measured Values
thetac = 291 deg
Tc = 2.35
%error
Error in thetac = 0.388%
Error in Tc = 3.422 %
Case 3:
Ta = 6.86
thetaa = 30.9
Tb = 5.488
Tc = 4.116
so from condition of static equilibrium
Ta(cos(30.9)i + sin(30.9)j) + 5.488(cos(thetab)i + sin(thetab)j) + 4.116(cos(thetac)i + sin(thetac)j) = 0
comparing
5.488cos(thetab) + 4.116cos(thetac) = -5.8863
5.488sin(thetab) + 4.116sin(thetac) = -3.522
(-5.8863 - 4.116COS(thetac))^2 + (-3.522 - 4.116sin(thetac))^2 = 5.488^2
34.648 + 16.9414coa^2(thetac) + 48.4560216cos(thetac) + 12.616704 + 16.941456sin^2(thetac) = 30.118144
34.648 + 16.9414cos^2(thetac) + 48.4560216cos(thetac) + 12.616704 + 16.941456sin^2(thetac) = 30.118144
solving
thetac = 267 deg
thetab = 177.2 deg
measured values
thetac = 269 deg
thetab = 179 deg
%error
error in thetac = 0.749 %
error in thetab = 1.015%
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