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)
RANGSIT UNIVERSITY College of Engineering
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
โมเมนตของเวกเตอร (moment of vector) คือนัยทั่วไปของแนวคิดโมเมนตของแรง โมเมนต M ของเวกเตอร B รอบจุด A หาไดจาก MO = rAB x B เมื่อ rAB คือเวกเตอรจากจุด A ไปยังจุดที่เวกเตอร B กระทํา และเครื่องหมาย x หมายถึงผลคูณไขวของเวกเตอร ดังนั้น M อาจกลาวไดวาเปน "โมเมนต M ที่เกี่ยวของกับแกนซึ่งวิ่งผานจุด A" หรือเรียกสั้น ๆ วา "โมเมนต M รอบจุด A" ถาจุด A เปนจุดกําเนิด (หรือบริบทที่เกี่ยวกับแกนหายไป) เราจึงสามารถ เรียกไดวาเปน โมเมนต เฉย ๆ และถา B เปนเวกเตอรของแรง โมเมนตของแรงก็จะเปน 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 กับแขนขอตอมุมฉาก ดังรูป
จงหาโมเมนตรอบจุด B
Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department
RANGSIT UNIVERSITY College of Engineering
Asst.Prof.Yannavut supichayanggoon Mechanical Engineering Department