0:00
Let's talk about this question.
00:01
We are given a figure wherein a post, where it shows a person, the weight of 6 .81 newton, so that's weight, remember not mass, doing push -ups.
00:11
So the weight exerts on cg, cg as in center of gravity, and this is the weight which is acting downwards.
00:18
We have to find a normal force which is exerted by the floor on the hand.
00:21
So this floor is exerting some normal reaction on the hand, which we have to find, we are given that the person holds this position so we can say that the person is at rest at the moment and we have to also find the force and we have to also remember that the force the floor also exerts a normal force on the foot as well so let's try to draw an fbd over here if we are given that this is w let's call this as the normal reaction n1 something which you have to find this is the normal reaction and two so first of this is an this is at risk so so we can say that it is in rectilinear, the rectilinear forces acting should be all zero.
01:07
So w should be equal to n1 plus n2.
01:11
So we can say that n1 plus n2 is equal to w and w is already given as 681.
01:17
So we're going to replace w by 681 and let's mark this as equation one.
01:21
Likewise we are going to say now that it is in rotational equilibrium as well.
01:26
So about the center of gravity about this point, the net torque should be zero...