This is when a vertical line passing through the centre of gravity of the body is on the point of falling outside of the body's base area. However if the plane is tilted, this state of affairs only remains until what is called the 'tipping point'. Normally the body sits on the plane in equilibrium, with the plane horizontal. This is to do with a rigid body in contact with a rough plane. Let the centroid be at position x C from the origin O. The radius r of each slice is given by Pythagoras' Theorem: The hemispherical axis is the x-axis and this time we consider circular slices of thickness δx. The method of calculation for the centre of mass of a 3D object is very similar to that of a 2D object. So t he centre of mass is on a horizontal line a third of the height up from the base. The constituent masses often take the form of thin slices( laminae) of the regular solid.Ĭonsider a small horizontal strip of area δA(delta A). When trying to calculate the centre of mass of a uniform solid eg a cone or hemishpere, we consider the whole mass and its moment about an axis and equate this to the sum of all the moments of constituent masses about the axis. The centre of mass (sometimes termed the centroid)
This effective weight may be considered to act through a particular point called: Assuming these forces are parallel, the weight of the body equals the sum of all the tiny particle weights. To understand the concept of centre of mass we must first consider a rigid body to be made of an infinite number of particles.Įach particle has a gravitational force on it directed towards the centre of the Earth. The turning effect of two equal and opposite parallel forces acting about a point equals the product of one force(F) and the perpendicular distance between the forces(d). When the beam is balanced, the sum of the downward forces equals the sum of the upward forces.Ī beam of negligible weight is horizontal in equilibrium, with forces acting upon it, as shown in the diagram.Ĭalculate the value of the weights R and W. One force acts vertically downwards near the centre, while the other acts vertically upwards at the end. If the reaction force at the fulcrum is R,Ĭonsider two forces F 1 and F 2 acting on a beam of negligible mass. horizontal), the sum of the downward forces equals the sum of the upward forces. When the beam is balanced at its fulcrum O(i.e. When attempting problems concerning a balance points or fulcrum, remember that there is always an upward 'reaction' force acting.Ĭonsider two forces F 1 and F 2 acting vertically downwards at either end of a beam of negligible mass. (the sum of clockwise & anti-clockwise moments = 0) Ii) there is no net turning effect produced by the forces I) the vector sum of the coplanar forces = 0 distance ( x)to the point of application.īy convention, anti-clockwise moments are positive.įor a rigid body acted upon by a system of coplanar forces, equilibrium is achieved when: The moment M (turning effect) of a force about a point O is the product of the magnitude of the force (F) and the perp.
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