The bent rod in fig 4 is supported at a by a journal bearing...

The bent rod in Fig: 4 is supported at A by a journal bearing: at D by ball-and-socket joint; and at B by means of cable BC: Using only one equilibrium equation, obtain direct solution for the tension in cable BC: The bearing at A is capable of exerting force components only in the z and y directions since it is properly aligned on the shaft: In other words, no couple moments are required at this support: 0.5 m Free-Body Diagram: As shown in Fig b. there are six unknowns Equations of Equilibrium: The cable tension TB may be obtained directly by summing moments about an axis that passes through points D and A: Why? 05 m 100 kg

Transcript

00:01 So here it is given that for a rod which is bended having the weight w will be equals to m into g so this will be equals to now mass of the rod is 100 kg multiplied by value of g is 9 .81 so it will comes out 981 newtons.

00:19 So in first part we have to draw the freeboard diagram.

00:23 So let's suppose this is what the bended rod according to quotient this point is given a this point is given b this point is given d.

00:33 So if we see the forces so at a point the vertical force will be along the z that is a of z while horizontal component at point a will be along the y axis that is a of y.

00:48 At this b point there will be a tension which will be in upward direction let's say tb.

00:55 If you remember so here is the center of mass for this bended rod so its weight will be exact a force in downward direction which would be equals to its weight that is w.

01:09 Now at point d there will be a vertical component of force which will be along the z direction that is dz while the horizontal component will be along the y direction that is dy.

01:24 Now if you see from d point to a point there will be a vector along this ad which will be the coefficient of friction that is mu ad while if you see this angle so this angle will be 45 degree.

01:45 So this is what the freeboard diagram.

01:47 Now in b part means in second part we have to find the equilibrium equation.

01:55 So for that we will use the newton's law along the y direction that is summation of fy this should be equals to zero.

02:03 So reaction at point a along the z will be r az plus similarly reaction at point d will be r dz plus tension at b will be tbc minus of the weight w this summation would be equals to zero...

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