Sample - Stat - Moment - 1
DERIVED EQUATION FOR THE DIAGRAM ABOVE
--> -0.809016994F1 + 0.615661475F2 = -257.1150439
--> -0.587785252F1 - 0.788010753F2 = -306.4177772
--> -0.809016994F1 + 0.615661475F2 = -257.1150439 EQ(1)
--> -0.587785252F1 - 0.788010753F2 = -306.4177772 EQ(2)
--> Solve for X in EQ(1)
--> -0.809016994F1 = -257.1150439 -0.615661475F1 EQ(3)
--> Divide the whole of EQ(3) by -0.809016994
--> F1 = ( -257.1150439 -0.615661475F2 ) / -0.809016994 EQ(3)
--> Substitute EQ(3) into EQ(2) to get F2
--> -0.587785252F1 -0.788010753F2 = -306.4177772 EQ(2)
--> -0.587785252 x ( -257.1150439 -0.615661475F2 ) / -0.809016994 -0.788010753F2 = -306.4177772 EQ(4)
--> Multiply the entire equation by the value of the denominator
--> -0.587785252 x ( -257.1150439 -0.615661475F2 ) + 0.637514090631736F2 = 247.897189018506 EQ(4)
--> 151.128430871753 + 0.361876735229567F2 0.637514090631736F2 = 247.897189018506 EQ(4)
--> 151.128430871753 + 0.999390825861303F2 = 247.897189018506
--> Conbine like terms together e.g the constant terms
--> 0.999390825861303F2 = 96.7687581467532
--> Solve for F2 by dividing both sides of the equation by 0.999390825861303
--> 0.999390825861303F2 / 0.999390825861303 = 96.7687581467532 / 0.999390825861303
--> F2 = 96.7687581467532 / 0.999390825861303
--> F2 = 96.8277431037604
--> F2 = 96.83
--> Solve for F1 by substituting the value of F1 into EQ(3)
--> F1 = ( -257.1150439 -0.615661475F2 ) / -0.809016994 EQ(3)
--> F1 = ( -257.1150439 -0.615661475 x 96.8277431037604 ) / -0.809016994 EQ(3)
--> F1 = ( -257.1150439 + -59.6131111401822 ) / -0.809016994 EQ(3)
--> F1 = ( -316.728155040182 ) / -0.809016994 EQ(3)
--> F1 = 391.497530199201
--> F1 = 391.50
--> F1 = 391.50 N
--> F1 = 96.83 N
2.54 Cm = 1 Inch
30.48 Cm = 1 Ft
100 Cm = 1 M
100 Cm = 3.28 Ft
Gravity(g) = 9.82 M\S^2 [ S.I Unit ]
Gravity(g) = 9.82 x 3.28
Gravity(g) = 32.22 Ft\S^2 [ Standard Unit ]
Sample - Stat - Moment - 1 - End
Sample - Stat - Moment - 2
First Case For The Given Diagram Above
NEGLECTING THE DIAGRAM WITHOUT ANY MASS OR FORCE AT M1
THE STATED \ GIVING EQUATION IS PROVEN \ STATED RIGHT
M1 --> 0 Kg = 0 N
M1 APPROACHES ZERO
G = 9.82 M\S^2
W16 = 9.82 x 16 = 157.12 N
W36.5 = 36.5 x 9.82 = 358.43 N
Sum Of The Forces In Y Direction Is Equal To Zero
G * M16 - G x M36.5 + R1 = 0
9.82 x 16 - 9.82 x 36.5 + R1 = 0
157.2 - 358.43 + R1 = 0
-201.31 + R1 = 0
R1 = 201.31
R1 = 201.31 N
The Force \ Reaction at R1 is positive and it is equal to = 201.31 N
A Negligible mass of 0 Kg is hunged \ positioned at M1 which resulted into a Force of 201.31 N
Second Case For The Given Diagram Above
A Mass Of Specific Weight \ Value Is Hunged \ Positioned At M1
Sum Of the Moment At R1 Which Is The Support IS Equal To Zero
Right Direction Is Positive ( ----> + )
-36 x M1 + 358.43 x 14 - 40 x 157.12 = 0
-36M1 + 5,018 - 6284.8 = 0
-36M1 - 1,266.8 = 0
-36M1 = 1,266.8
M1 = 1,266.8 / -36
M1 = -35.19
M1 = -39.19 N
The Force At M1 = -39.19 N
Force M1 = 39.19 N (Down)
The Actual Mass At M1 = -39.19 / 9.82
Mass(M) = 3.58 Kg
Sum Of the Forces In The Y Direction IS Equal To Zero
157.12 - 358.43 + R1 - 35.19 = 0
R1 - 236.5 = 0
R1 = 236.5
R1 = 236.5 N
When A Mass Of 3.58 Kg Is At M1, The Vertical \ Reaction At R1 = 236.5 N
Sample - Stat - Moment - 2 - End
