Q on forces in equilibrium.

[Sonal7]

I am stuck on part (c). I have uploaded the answer to the first two parts. I cant understand the answer to part c. Why does the resultant force go through M. I dont understand why there is no vertical component. Thank you.

 

[Deleted member 4993]

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I am stuck on part (c). I have uploaded the answer to the first two parts. I cant understand the answer to part c. Why does the resultant force go through M. I dont understand why there is no vertical component. Thank you.



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The wording of the problem is very peculiar. It states:

The pole is smoothly hinged at A - that should mean that at A there is no moment reaction and it can have reactions in the x and y directions (A_x & A_y)​

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Then it states - end C of the of rod CD is fixed - however the drawing is indicating that CD is a two-force member with no "moment reaction" at C. That can only happen if joint C is hinged (not fixed).​

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Then it states end D of CD is "freely jointed" - what does that mean regarding reactions at D? I don't know!​

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What is the text-book that you are following for these problems. May be you search the index (or the glossary) to clarify these terms.​

 

[Sonal7]

The wording of the problem is very peculiar. It states:

The pole is smoothly hinged at A - that should mean that at A there is no moment reaction and it can have reactions in the x and y directions (A_x & A_y)​

​

Then it states - end C of the of rod CD is fixed - however the drawing is indicating that CD is a two-force member with no "moment reaction" at C. That can only happen if joint C is hinged (not fixed).​

​

Then it states end D of CD is "freely jointed" - what does that mean regarding reactions at D? I don't know!​

​

What is the text-book that you are following for these problems. May be you search the index (or the glossary) to clarify these terms.​

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Well some of it makes sense. Like smooth means friction free, so you ignore any friction. So there is just the reaction force at A to consider. I got the rest but unsure about c. I think in an exam they would have to make it super clear what is being asked.

 

[Deleted member 4993]

Well some of it makes sense. Like smooth means friction free, so you ignore any friction. So there is just the reaction force at A to consider. I got the rest but unsure about c. I think in an exam they would have to make it super clear what is being asked.

My point is - all of those joints (3) could be smooth joints. By using 3 different terms, they actually made the problem "statically indeterminate" - impossible to solve with only statics.

 

[Sonal7]

My point is - all of those joints (3) could be smooth joints. By using 3 different terms, they actually made the problem "statically indeterminate" - impossible to solve with only statics.

someone solve it. The three forces must pass through the same point for the particle to be equilibrium, otherwise there be a turning force. The moment must add up to zero. THank you however.

 

[Deleted member 4993]

someone solve it. The three forces must pass through the same point for the particle to be equilibrium, otherwise there be a turning force. The moment must add up to zero. THank you however.

Yes... but the reaction force at M due to the hinge at M - will have a component perpendicular to CM. They have tactfully ignored that!