Three chitdren are standing on a 15 ×15-ft raft. If the weigta of the children at points A, B, a

Three chitdren are standing on a 15 ×15-ft raft. If the...

Key Concepts

Resultant Force

The resultant force is the single force obtained by vectorially adding multiple individual forces. It represents both the net magnitude and the overall direction of the forces acting on a system, and in the context of weights on a body, it is found by summing the magnitudes of the forces and deducing their combined effect on the point of application.

Center of Mass (Center of Gravity)

The center of mass (or center of gravity) is the point where the total weight of a system can be considered to act. Determining it involves taking a weighted average of the positions of individual masses and is crucial for understanding balance and stability in static systems.

Moments of Force

Moments of force measure the tendency of a force to cause rotation about a point. In problems involving multiple weights, calculating the moments about a reference point helps in determining the position at which the resultant force effectively acts to maintain equilibrium.

Transcript

00:01 Problem 3 .127.

00:02 In this problem, we have to find magnitude of equivalent force and point of application of the force because of these three weights acting on the system.

00:14 So first of all, let us make a little bit effort to draw a rough sketch of equivalent diagram.

00:21 So let's say this was the stand.

00:24 All right and this was y -axis this was z -axis and this is x -axis and this is x -axis and let us say that our equivalent portion of magnitude r is acting in x -z plane and its coordinate since it is in x -z plane so its coordinate will be x comma 0 -z so our first effort will be to calculate calculate the value of r.

00:57 So for that, we can balance the forces along y direction.

01:01 So if you balance the forces along y direction, so r comes out to be this 375 newton is in negative y direction, so this will be minus 375 newton j cap.

01:14 This will be minus 260 newton j cap.

01:19 And the third force is again in the negative y cap, so in the negative y direction, so it will be minus 400 newton j cap so if we add them it comes out to be minus 1 ,035 nukin j cap now in the problem they are asking to calculate magnitude of this resultant force so magnitude will be thousand thirty five newton so this is the magnitude now they are asking to find its point of application now we have already assumed its coordinate to be x .0 comma z so to calculate x x is the distance from z x so this distance is going to be x for that we can balance moment about z xx's so if we balance moment about x xx's so in the equivalent diagram the moment because of this r will be 1 ,035 multiplied by perpendicular distance and this should be equal to the the moment because of the individual forces about the xxies so we can add up the moments because of individual forces so this 375 newton is at a distance of this distance is given as 1 meter so multiplied by 1 this 260 newton is at a distance of 1 .5 meter so we can write plus 260 multiplied by 1 .5 and the third one now this distance is given as 0 .25 meter and this entire distance since this was 5 by 5 box so this entire length is 5 meter and this length is 0 .25 so length so the distance from the of point c from the x x will be 5 minus 0 .25 so this will be come 400 multiplied by 5 minus 0 .25 and if we talk about its unit so all the units are in newton meter so we can get the value of x from this expression and x comes out to be 2 .5749 so we have got the x coordinate of this point of application now we have to calculate z coordinate and z is the direction along z -axis so this distance will be from x -axis so this direction is denoting here z so for that we can balance moment about x -axis so if you balance moment about x -axis so we can write as moment about x -axis will be if we observe in this uh in this equivalent diagram so the moment value about x -axis will be are multiplied by z and that will be equal to a moment, total moment, about x -axis because of all these three forces acting in this real diagram.

04:38 So if we talk about the distance from x -axis of point a, so this distance, since this entire side was of 5 meter and this length is 2 meter, so distance from x -axis will be 3 meter...

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