Phys2425 LAB REPORT FORM EQUILIBRIUM AND VECTORS: THE FORCE TABLE Data Table I:

Question

Phys2425 LAB REPORT FORM EQUILIBRIUM AND VECTORS: THE FORCE TABLE Data Table I: Determination of Equilibrant Forces: E, F, G, and S M (kg) Equilibrant M - 9.8 (N) (Force) 0 (Degree) (Direction) (Mass) 0. A0 kg O, 190 kg O, 2012 1886 . 9678 Data Table 2: Applied Forces M (kg) M. 98 (N) (Force) O (Degree) (Direction) Applicd Forces (MAS) 0 I 10 1.27 370 O. 105 1.03 0.105 3230 103 90° O 54 0.055 oo 0,105 1.03 so 0.068 2400 O.I 30 1.27 Data Table 3: Graphical Method to Determine the Equilibrant E for A + B +E-0 Graphical Result: E - kg deg Phys2425 Data Table 4: Graphical Method to Determine the Equilibrant S for P + Q +R+S-0 Graphical Result: S = kg 0 = deg

Explanation / Answer

Given

vectors are

A = 1.27 N at an angle of 37 degrees

B = 1.03 N at an angle of 143 degrees

-B = 1.03 N at an angle of 323 degrees

P = 1.03 N at an angle of 0 degrees

Q = 0.67 N at an angle of 50 degrees

R = 1.27 N at an angle of 240 degrees

Data table 3

A+B+E =0

E = -(A+B)

E = -A -B

vector components are  

A = 1.27 cos0 i + 1.27 sin0 i = 1.27 i + 0 j

B = 1.03 cos143 i + 1.03 sin 143 i =-0.823 i + 0.62 j

-B = 1.03 cos323 i + 1.03 sin 323 i =-0.62 i +0.823 j

P = 1.03 cos0 i + 1.03 sin 0 i = 1.03 i + 0 j

Q = 0.67 cos50 i + 0.67 sin50 i = 0.43 i + 0.51 j

R = 1.27 cos240 i + 1.27 sin240 i = -0.635 i -1.1 j

now E = -A-B = - 1.27 i + 0 j -0.62 i +0.823 j= (-1.27-0.62)i +0.823 j = -1.89 i +0.823 j

the magnitude is E = sqrt( (-1.89)^2+0.823^2) N = 2.06 N

the direction is theta = arc tan ( 0.832/(-1.89))= -23.76 degrees from the x axis

that is below the x axis  

P+Q+R+S = 0  

S = -(P+Q+R)

S = - (1.03+0.43-0.635)i +(0+0.51-1.1) j = -(0.825 N i -0.59 N j ) = -0.825 N i + 0.59 N j

the magnitude is S = sqrt( (-0.825)^2+(0.59^2)) N = 1.014 N

the direction is theta = arc tan (0.59/(-0.825))= -35.57 degrees

we can draw the graph accordingly

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