Test: Centroid Of A Body - Civil Engineering (CE) MCQ

The density is the ratio of the mass to the volume of the body. Thus arranging them and the volume is said to be as ratio of mass to density. This equation is generally used in the calculations of the centroid of the body.

The answer is obviously no. Still if we are having any difficulty in making the vector components, then we can go in 2D. As if the particle is in equilibrium, the net force will be zero. No matter where you see first. Net force is zero. But the statement here is not correct about the centroid.

In the simplification of the loading system the net force acts at the centroid of the loading body. That is if the loading system is in the form of the triangle then the at the distance 2 by 3 of the base the net force of the loading will act. And the load will be half the area of the loading.

The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force. Thus the use of centroid.

The external forces are treated differently. They are not added by the force of the distributed loading. That is the force not only acts at the centroid always. It can be shifted also. Depending on the external forces. Thus the use of centroid.

The calculation is general to all the axis of coordinate system. This means that the calculations are generalised for all the axis. All the axis are given same priority. Thus no change in calculations.

The product of the centroid of the section to its volume to the total volume of the body is the centre of volume. Thus the answer. The ratio is generally used to locate the coordinate of the centroid or the centre of volume.

It is given that x, y and z-axis components are Z, X and Y respectively. Thus, just by placing the right coordinates we get option d to be correct. Here just the interchange of the axis’s representor with their respective axis is creating confusion. Which must be figured out.

As given the vector is F = Ai + Bj + Ck, this implies that the x ,y and z-axis components of this vector is A, B and C respectively. But, in rectangular components representation of any vector, the vector is written as F = Fx + Fy + Fz.

The weight of the body is the sum of the all the small weights that are been applied by all the particles. The smaller the particle the smaller the weight it applies. Thus the total weight of the body is the summation of all the infinite number of small weights applied to the body.

The force developed by a support doesn’t allow the translation of its attached member. This is the basic condition for the equilibrium of the forces in any dimension. This rule is applied when the support reactions are taken into the account for the equilibrium of the body.

For the equilibrium in the three dimensional system of axis we have all the conditions true as, ∑Fx=0, ∑Fy=0 and ∑Fz=0. Also we have the summation of the forces equal to zero. Which is not a non-zero value.

As 3D is defined as the three axis system, we have to consider the equilibrium in all the three axis. This will make the equilibrium go on all the axis of the 3D space. And hence will cancel all the forces. These are the points to be kept in considerence while determining the centroid.

Yes, the equilibrium is being achieved w.r.t the ground. Like the motion w.r.t ground need be zero. That is the relative velocity of the object or the body must be zero w.r.t the ground. This means motion is in equilibrium. This equilibrium is also defined by the centroid axis.

The weight of the body is the sum of the all the small weights that are been applied by all the particles. The smaller the particle the smaller the weight it applies. Thus the total weight of the body is the summation of all the infinite number of small weights applied to the body.