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Lab 103: Translational Static Equilibrium-Force Table Objectives 1. To confirm t

ID: 3280548 • Letter: L

Question

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 111A

Explanation / 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%