# Replace the force system by an equivalent resultant force and couple moment at point o

Replace
the force system by an equivalent resultant force and couple moment
at point O. Take F3 = -200i + 500j -300k

General guidance

Concepts and reason
The concepts used in solving this problem are vector algebra, force, and moment about a point.
Vectors:
Vectors are those quantities which have both magnitude and direction. Such as torque, moment, acceleration and force.
Vector algebra:
This is a bunch of concepts which describes the methodology of different vector operations, such as vector product and scalar product of vectors.
Position vector: Position vector shows the location of a point in a vector space in terms of the unit vectors in different directions.
Moment about a point:
Moment is the turning effect of force and is usually defined with respect to a fixed reference point. The evaluation of moment is done by multiplying the magnitude of the force and the perpendicular distance of force from a reference point.
Initially, draw the diagram of the given system. Next, write the forces in cartesian form and determine the resultant force. Next, determine the position vector for the point of action of the forces from point O in cartesian form. Finally, determine the moment about point O by performing vector multiplication of forces with their respective position vectors.

Fundamentals

The expression of moment is written as,

Here, is the moment, is the perpendicular distance, and is the force.
The expression for moment about point for the below arrangement is written as,

Here, is the net moment about point , is the force at point B, is the position vector of point B from C, is the force at point A, and is the position vector of point B from A.
A position vector, for a point A, with respect to origin is represented in Cartesian form as,

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Here, , , and are the distances of point A from the origin in x, y, and z directions respectively.
Vector multiplication:
For two vectors written as,

Here, , , and are the x, y, and z components of respectively and , , and are the x, y, and z components of .
Vector multiplication or cross product for two three-dimensional vectors is,

Step-by-step

Step 1 of 2

Draw the diagram of the given system.

Here, , , and are the forces applied on the system.
Write the expression for in the cartesian form.

Substitute for .

Write the expression for in the cartesian form as provided in the question.

Substitute for .

Write the expression for in the cartesian form.

Write the expression of the resultant force on point O.

Substitute for , for , and for .

The resultant force is .

The resultant force applied on point O is the sum of all the forces applied on different points. Since the forces act in different direction the resultant force is in vector form. Thus, firstly converting all the forces into their cartesian form and then determining the resultant.
The direction of force is along the negative z axis, so its vector form is written as . The direction of force is along the positive y axis, so its vector form is written as .

Incorrect substitution will lead to wrong answer. Always write the substitution statement before substituting the values of known parameters.

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Determine the position vectors for the point of action of the forces from point O in cartesian form. Determine the moment about point O from the expression of moment.

Step 2 of 2

Write the expression for the position vector between and point O in the cartesian form.

Here, is the distance between and point O and is the position vector of point of action of from point O in the cartesian form.
Substitute for .

Write the expression for the position vector between and point O in the cartesian form.

Here, and are the distance between and point O in x and y directions and is the position vector of point of action of from point O in the cartesian form.
Substitute for and for .

Write the expression for the position vector between and point O in the cartesian form.

Here, and are the distance between and point O in x and y directions and is the position vector of point of action of from point O in the cartesian form.
Substitute for and for .

Write the expression of moment about point O.

Substitute for , for , for , for , for , and for .

The couple moment about point O is .

To calculate the couple moment about a point initially write the distance between the force applied and point O in cartesian form. The point of action for force is on the positive y axis, so its position vector is . The point of action for force and is on the x-y plane, so their position vectors are and respectively.
Substitute the value of force and the distance in the expression of moment about a point. To calculate the resultant moment about a point, add all the moments acting at that point.

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Do not use for the expression of moment. Always use as the expression of moment.

The resultant force is .

The couple moment about point O is .

The resultant force is .

The couple moment about point O is .

M=rxF
.
Mc = (‘mcXF )+(racxF;)
ro=(aji+a,j+a,k)
A=(cî+cj+ck) B= (b + +byk)
Ах B = a a, a ь ь, ь,
2 m F1 = 300 N 1 m TL 1.5 m 1.5 m = 200 N
F =-FK
300 N
F =-300 N
F, = FÎ
200 N
F, = 200ÎN
F} = (-200î +5009 – 300k) N
F=F+F,+F
-300 N
2009 N
(-200î +500 j – 300k N
F=-300Ê N+200Î N +(-200î +5009–300€) N = (-200î +700 9 – 600) N
rio=noj
120 = P20 i +120,
Mo=(1.5ỉ +3.5j) m
150 =150, +rso,
150 = (1.5ỉ +2ị) m
M. =(roxF)+(120xF_)+(130xF)
-300 N
2009 N
(-200î +500 j – 300k N
(1.5i +2j) m
((2m)x ((1.sî +3.5ý) m)x).(((1.5i +2j) m)x ***|(-3006 N))*((2009 N) ‘lll-200 + 5009–300k) N)) м. = 0 2 0 +1.5 3.5 0+ 1.5 2 0 10 0 -3000 2000-200 500 -300 = (-600î)+(300k)+(-600ỉ + 450þ +1150k) = (-1200î +4509+1450h) N-m
rio=noj
120 = P20 i +120,
150 =150, +rso,
M=Fxr
M=rxF
(-200î + 7009 – 600) N
(-200î + 700ſ – 600k) N
(200î +700ſ +600k) N
(-600î +2009 – 700k) N
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