Question 10 points the pipe abc shown in the figure below is...

Question: [10 points]: The pipe ABC shown in the figure below is supported by three radial bearings at A, B and C and is subjected to the 450 N external force acting as shown in the figure. Neglect the thickness and mass of the pipe and assume that the radial bearings at A, B, C have zero moments about the lines normal to the shaft. a) Draw the FBD of the pipe ABC onto the diagram below. (3) b) Determine the reactions at A, B and C. (7) [hint: a radial bearing supports only radial load i.e. no reaction force along the shaft axis]

Transcript

00:02 In the first part of the question we have to replace the supports at point a and b with their reaction forces.

00:11 Here it will be r a, or let's say rxa and there will be rya and here there will be r b and here there will be only rb or ryb rxb will be equals to 0 because no reaction force will be active in x direction because there is a roller attached here so any force acting will simply move the whole plank or whole beam.

00:48 Now coming to the fbd, so here it will be rxa, here it will be rya, here it will be rya, here it will be ryb and if you take the component of this force that is 450 newton, that is, that is it is 30 degrees so if you take the component vertically it will be 450 sign 30 so it will become equals to 225 newton and horizontally it will be equals to 450 equals to 450 into cost 30 so that equals to 389 newton and here at rb there is another force acting that is equal to 300 newton this is point a this is point b so this is the fbd for the beam now coming to the writing the equilibrium equation one can see here balancing the force in x direction that is in horizontal direction using the concept of newton's law of motion one can say that rya plus this force 389 newton must be equals to 0 so that overall beam is in equilibrium position or equilibrium condition while examining the in horizontal direction sorry here i did the mistake here it will be rya and here it will be rxa so one can write rxa plus 389 newton equals to 0 so from here we get rxa equals to minus 389 newton.

02:50 Now coming to the other part of the question or finding this force, let's say, ryb, let's say we take moment from point a.

03:07 Then when taking moment from point a and writing the product of all the moment acting on this beam, sorry some of all the moment acting on the beam we can have 225 into 2 meter plus 300 into 6 meter here whole length is 6 meter and here this length is 2 meter minus this all moment is counter by ryb so it will be ryb into 6 meter equals to 0 so from here we can calculate ryb equals to, so that comes equals to 375 newton.

03:54 So this force will be equals to 375 newton...

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