Coplanar Non-Concurrent Forces – Problem 1 – Resolution and Composition of Forces
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Coplanar Non-Concurrent Forces – Problem 1 – Resolution and Composition of Forces

October 23, 2019


here are some problems based on the concept of moments and here I have written that these are problems on co-planner non-concurrent for system we would be solving question number one see I’m marking question number one let us read what is given here a force of 1200 Newton X on a bracket as shown in Figure here in this question they have given as the figure diagram is provided to us here there is a bracket which we can see in this particular shape this is a bracket this bracket is having point a which we can say is fixed to a wall here there is a wall over which the bracket is fixed the length of this bracket from point A up to the vertical height that is 140 mm and this vertical portion is 1200 mm over this vertical portion there is a load of 1200 Newton inclined at 30 degree with the horizontal here we have to find the moment of this force that is 1200 Newton about point a which is fixed to the bracket now let us see how we can solve this problem question number one 1 I will write the solution for this first by drawing a diagram which they have given of a bracket here there is a bracket at this vertical portion whose height is 1200 mm we are having a load of 1200 kilo Newton or you can say 1200 Newton it has given 1200 not kilonewton inclined at 30 degree next the length of this bracket is 1 4 0 140 mm bracket is attached to the wall at point a at Point a we need to calculate the moment of this force 1200 Newton about point a so here the question is calculation of moment about point a how to get this how to calculate the moment first of all see that the force which they have provided since it is inclined here you will have two this of wait you cannot solve the problem until and unless you have resolved the force which is an inclined force if I can see here this 1200 Newton is inclined with horizontal at 30 degree so here I will have two components one is horizontal and the other is vertical horizontal component I can write it as force into cos of angle which is 30 degree vertical component I can write it as force into sine of angle which is 30 degree now I have to find the moment of these two components about point a because we cannot find the moment until and unless we convert the incline force into its components you cannot find the moment so first of all you have to convert the incline force into a two of its component that is horizontal and vertical component I hope you are understanding this now after this I will say that taking moments of components of 1200 Newton force about point a I will be taking the moments of components of 1200 Newton force above point a so I will I will be taking these two components now we have to understand how to take the moment I will consider 1200 sine 30 first very home look a force taking first we will see 1200 sine 30 now I have to take the moment of this force about point a so if I see 1200 sine 30 its location with respect to point a the distance is 140 so the moment I want to bring this 1200 sine 30 towards point a so it means in short I have to take the moment in anti-clockwise direction then only it will be reaching point a so I will say that therefore moment at a will be equal to anti-clockwise moment we will consider it as negative clockwise moment we will be considering it as positive so 1200 sine 30 is anti-clockwise so it will be minus 1200 sine 30 into as per the definition of moment it is forced into perpenicular distance force is 1200 sine 30 it’s perpendicular distance with point a is 140 this is the moment of 1200 sine 30 about point A next I have another force which is 1200 cos 30 now I will have to take 1200 cos 30 at Point a if I am taking this here we would be getting a moment in clockwise direction at Point a so this will be plus 1200 cos 30 into 1200 so if I calculate it carefully all the values I will be getting my answer of moment which I have
103
00:07:08,610 –>00:07:18,890
here it is 40.71 forty point seven one into 10 raised to 3 Newton meter 10 raised to 3 Newton meter since the answer is positive I can say that the total effect will be in the clockwise direction and with this we have solved the question here only the moment of a force was asked

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  1. sir while resolving shouldn't we take cos theta as the vertical component? then why have you took sin theta as the vertical component

  2. Hey guys, The correct answer is 1163.076N-m because he made a mistake on his calculation he have used 1400mm instead of 140mm. if you used 1400 then the answer will be 407076 which is wrong. Hope this help you

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