State the conditions for the equilibrium of a rigid body.
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Equilibrium of a Rigid body
A rigid body is said to be in mechanical equilibrium, if both its linear momentumand angular momentum are not changing with time. In other words, the body is in mechanical equilibrium when it has neither linear acceleration nor angular acceleration.This means the conditions for equilibrium can be expressed as the following:
conditions for equilibrium
(1) The total or net force i.e. the vector sum of all the forces, on the rigid body is zero.
i.e., ΣFi = 0 when i=1 to n …………………… (1 a)
or
dptot/dt = 0 where ptot=net linear momentum………………… (1b)
(2) The total Torque i.e. the vector sum of the torques on the rigid body is zero.
ΣΤi = 0 when i=1 to n………………….. (2a)
or
dltot/dt = 0 where ltot=net angular momentum…………………(2b)
Conditions for mechanical equilibrium
Expanding equation 1a and 2a:
ΣFix = 0 ΣFiy = 0 ΣFiz = 0 when i=1 to n
ΣΤix = 0 ΣΤiy = 0 ΣΤiz = 0 when i=1 to n
For m.e. these 6 independent conditions are to be satisfied.
Conditions when forces are coplanar
We need only 3 conditions to be satisfied for this when the forces working on the rigid body are coplanar.
* Two of these conditions correspond to translational equilibrium: The sum of forces along any 2 perpendicular axes in the plane must be zero. so if we take x and y axes, then ΣFx = 0 ΣFy = 0
* The third condition is about Rotational Equilibrium: The sum of the torques along any axis perpendicular to the plane of the above said forces, must be zero. Here it will be: ΣΤz = 0
A rigid body is said to be in mechanical equilibrium, if both its linear momentumand angular momentum are not changing with time. In other words, the body is in mechanical equilibrium when it has neither linear acceleration nor angular acceleration.This means the conditions for equilibrium can be expressed as the following:
conditions for equilibrium
(1) The total or net force i.e. the vector sum of all the forces, on the rigid body is zero.
i.e., ΣFi = 0 when i=1 to n …………………… (1 a)
or
dptot/dt = 0 where ptot=net linear momentum………………… (1b)
(2) The total Torque i.e. the vector sum of the torques on the rigid body is zero.
ΣΤi = 0 when i=1 to n………………….. (2a)
or
dltot/dt = 0 where ltot=net angular momentum…………………(2b)
Conditions for mechanical equilibrium
Expanding equation 1a and 2a:
ΣFix = 0 ΣFiy = 0 ΣFiz = 0 when i=1 to n
ΣΤix = 0 ΣΤiy = 0 ΣΤiz = 0 when i=1 to n
For m.e. these 6 independent conditions are to be satisfied.
Conditions when forces are coplanar
We need only 3 conditions to be satisfied for this when the forces working on the rigid body are coplanar.
* Two of these conditions correspond to translational equilibrium: The sum of forces along any 2 perpendicular axes in the plane must be zero. so if we take x and y axes, then ΣFx = 0 ΣFy = 0
* The third condition is about Rotational Equilibrium: The sum of the torques along any axis perpendicular to the plane of the above said forces, must be zero. Here it will be: ΣΤz = 0
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When Newton's first law of motion is applicable, the rigid body is in equilibrium. An object can be in equilibrium when all the other forces are balanced properly. When such a situation arises, the net result of all the forces applied is zero.
The conditions for the equilibrium of a rigid body are as follows-
- As discussed above, the net result or vector sum of all the forces being exerted on the rigid object must be zero.
- The total torque or the sum of torques on the rigid body should also be zero.
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