CHAP03_RIGID_BODIES_EQUIVALENT_FORCE_SYSTEMS

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RANGSIT UNIVERSITY College of Engineering

Chapter 3 RIGID BODIES: EQUIVALENT FORCE SYSTEMS F The forces acting on a rigid body can be F’ separated into two groups: (1) external forces (representing the action of other rigid bodies on the rigid body under consideration) and (2) internal forces (representing the forces which hold together particles forming the rigid body. Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

F F’

According to the principal of transmissibility, the effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. Two forces acting on the rigid body at two different points have the same effect on that body if they have the same magnitude, same direction, and same line of action. Two such forces are said to be equivalent. Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

V=PxQ

The vector product of two vectors is defined as

Q θ

V=PxQ

P

The vector product of P and Q forms a vector which is perpendicular to both P and Q, of magnitude

V = PQ sin θ This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through θ which brings vector P in line with vector Q. The three vectors P, Q, and V - taken in that order - form a right-hand triad. It follows that

Q x P = - (P x Q)


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j k

i

It follows from the definition of the vector product of two vectors that the vector products of unit vectors i, j, and k are ixi=jxj=kxk=0

ixj=k , jxk=i, kxi=j , ixk=-j , jxi=-k, kxj=-i The rectangular components of the vector product V of two vectors P and Q are determined as follows: Given

P = Px i + Py j + Pz k

Q = Qx i + Qy j + Qz k

The determinant containing each component of P and Q is expanded to define the vector V, as well as its scalar components Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

P = Px i + Py j + Pz k i V = P x Q = Px Qx

Q = Qx i + Qy j + Qz k

j k Py Pz = Vx i + Vy j + Vz k Q y Qz

where

Vx = Py Qz - Pz Qy V y = P z Qx - P x Qz Vz = Px Qy - Py Qx Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

The moment of force F about point O is defined as the vector product

Mo

F

MO = r x F

where r is the position vector drawn from point O r O θ to the point of application d A of the force F. The angle between the lines of action of r and F is θ. The magnitude of the moment of F about O can be expressed as

MO = rF sin θ = Fd where d is the perpendicular distance from O to the line of action of F.


RANGSIT UNIVERSITY College of Engineering

โมเมนตของเวกเตอร (moment of vector) คือนัยทั่วไปของแนวคิดโมเมนตของแรง โมเมนต M ของเวกเตอร B รอบจุด A หาไดจาก MO = rAB x B เมื่อ rAB คือเวกเตอรจากจุด A ไปยังจุดที่เวกเตอร B กระทํา และเครื่องหมาย x หมายถึงผลคูณไขวของเวกเตอร ดังนั้น M อาจกลาวไดวาเปน "โมเมนต M ที่เกี่ยวของกับแกนซึ่งวิ่งผานจุด A" หรือเรียกสั้น ๆ วา "โมเมนต M รอบจุด A" ถาจุด A เปนจุดกําเนิด (หรือบริบทที่เกี่ยวกับแกนหายไป) เราจึงสามารถ เรียกไดวาเปน โมเมนต เฉย ๆ และถา B เปนเวกเตอรของแรง โมเมนตของแรงก็จะเปน แรงบิด Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


y

A (x , y, z )

Fy j yj Fx i

r O

Fz k

xi x

zk z

The rectangular components of the moment Mo of a force F are determined by expanding the determinant of r x F.

i j k Mo = r x F = x y z = Mx i + My j + Mzk Fx Fy Fz where

Mx = y Fz - z Fy Mz = x F y - y F x

My = zFx - x Fz


y

A (x A, yA, z A)

Fy j B (x B, yB, z B)

Fx i

r Fz k O

x

z

In the more general case of the moment about an arbitrary point B of a force F applied at A, we have

i

j

where

MB = rA/B x F = xA/B yA/B Fx Fy rA/B = xA/B i + yA/B j + zA/B k

and

xA/B = xA- xB

yA/B = yA- yB

k zA/B Fz

zA/B = zA- zB


y

F

Fy j rA/B

(yA - yB ) j

A

Fx i

B (xA - xB ) i

O z

MB = MB k

x

In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalar

MB = (xA- xB )Fy + (yA- yB ) Fx The right-hand rule is useful for defining the direction of the moment as either into or out of the plane (positive or negative k direction).


RANGSIT UNIVERSITY College of Engineering

แรงขนาด 800 N กระทํา กับแขนขอตอมุมฉาก ดังรูป

จงหาโมเมนตรอบจุด B

Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


RANGSIT UNIVERSITY College of Engineering

Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department


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