A uniform rod AB, 100cm long and of mass 2kg is supported horizontally at points...

A uniform rod AB, 100cm long and of mass 2kg is supported horizontally at points P and Q, 40cm and 65cm respectively from A. A mass of 1kg is hung at a point 25cm from A. If the rod is in equilibrium, calculate the reactions at the support(Take g= 10m/s²).?

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Answers (23)

EmX
3 weeks ago
Sorry for the delay.

As we already know, when dealing with moments, it's always about the forces about a pivot.

And in the diagram below, The two arrows at P&Q represent the reactions at those pivots.

Therefore it is the value of P& Q we're looking for in this question.


Before we go any further we know in equilibrium
Sum of Upward forces equals sum of downward forces

In this case the upward forces are P &Q.

The downward forces are the mass of the ruler (2kg)and the 1kg mass attached at 25cm from A.

Observe below
  • Robert Endurance: Thank you.
    Like 0    Dislike 0   3 weeks ago
  • EmX: Please notice that all the values are masses, but our answer is going to be in Newton since the reaction at each pivot is a force.

    Therefore we are going to convert all the masses in this question to forces by multiplying by 10 (f=ma).


    After that we take the sum of moments about each point using our second condition of equilibrium.


    Which is sum of Clockwise moments= sum of anticlockwise moments.
    Like 1    Dislike 0   3 weeks ago
  • EmX: Since we've already found Q, we can easily subtitute it into equation.....(1). And we'll able to find P.

    But that's for you to do If you want save time

    Let me do it the hard way, I'll use that equation...(1) to proove that my final answer Is correct.
    Like 0    Dislike 0   3 weeks ago
  • EmX: Now let's proove it using the equation...(1)
    Like 1    Dislike 0   3 weeks ago
  • EmX: Please if you have any confusions or further inquires on this question.

    Feel free to comment.
    I'll explain...to the best of my abilities
    Like 1    Dislike 0   3 weeks ago
  • Robert Endurance: No, I am good. Thank you for being helpful.
    Like 1    Dislike 0   3 weeks ago
  • Robert Endurance: Another, calculate the maximum mass that can be hung at a point 90cm from A.
    Like 1    Dislike 0   3 weeks ago
  • EmX: Sorry, I just came online.

    But if, the system is already in equilibrium before...the mass comes mass comes in, the mass has to be weightless to not disturb the equilibrium of the system.


    But If it's the new mass that brings about equilibrium, then I must ask.


    Are you referring to the same diagram of the last question.?
    Like 1    Dislike 0   3 weeks ago
  • Robert Endurance: Yes, the same diagram of the last question.
    Like 0    Dislike 0   3 weeks ago
  • EmX: If that's the case, then the mass that'll be added is zero.

    This is because the system is already at equilibrium in this question. Any additional mass will disrupt the equilibrium.

    If you doubt me.
    I can solve it for you. Just to proove that the answer will be zero.


    First of all, as usual the total force acting downward has to be equal to that acting upwards,

    After that, we take the moment around P...
    Like 1    Dislike 0   3 weeks ago
  • EmX: After that we clearly take the moments around Q.
    Like 1    Dislike 0   3 weeks ago
  • EmX: Now that we have the values of P and Q. We can now substitute for P and Q. In equation....(1).
    Like 1    Dislike 0   3 weeks ago
  • EmX: So as you can see. What ever mass you place anywhere will ruin the equilibrium.

    Maybe the only way you'd have gotten a reasonable value as an answer for your question would have been: if you changed the position of one of the pivots. (Like adjusted it a bit.).


    Sorry, I forgot to mention this at the beginning. But incase you don't understand.

    "W" that I keep using is the weight of the additional mass.
    Like 1    Dislike 0   3 weeks ago
  • Robert Endurance: Thank you for your assistance.

    I understand but can I know why you used 90w instead of (90 - 40)w and (90 - 65)w about P and Q respectively in the calculation above?

    I am looking forward to your swift and commendable reply.
    Like 0    Dislike 0   3 weeks ago
  • EmX: I'm so sorry about that.
    It's a mistake on my part. I didn't resketch the diagram and input the new value.

    I'm so sorry.
    Like 0    Dislike 0   3 weeks ago
  • Robert Endurance: Okay, thank you for you acknowledgement.
    Like 0    Dislike 0   3 weeks ago
  • EmX: Thanks for being observant.
    Again, I'm terribly sorry for the mistake.

    But it doesn't change the results
    Like 1    Dislike 0   3 weeks ago
  • EmX:
    Like 0    Dislike 0   3 weeks ago
  • EmX: Sorry, I'm not exactly feeling fine, and I cant quite think straight.
    so there's a possibility that I made another mistake.
    But I'm glad you get the idea.
    Like 1    Dislike 0   3 weeks ago
  • Robert Endurance: I express empathy with you. But take this quick question, is it possible for the answer to be a 1.5kg without considering the reactions, the pivot on the left and the mass 1kg?

    However, if you have no answer to this, you explanation above is brilliant.

    Thank you for your help.
    Like 0    Dislike 0   3 weeks ago
  • EmX: Let me get this straight. You want to remove the pivot at the left??
    Like 0    Dislike 0   3 weeks ago
  • Robert Endurance: Possibly not, but when I playfully solved, without considering the pivot on the left, the reactions at P and Q; the mass 1kg, I arrived at an answer quite surprising that matches with a textbook answer.

    This was unusual because it appeared that the author of Further Mathematics Project 3, disregarded the effects of other parameters but focused on the pivot on the right, the mass of the metre rule, 2kg and the maximum mass that could be hung at the 90cm mark.

    However, once again, your explanation on equilibrium was clear and helpful. And there is no need for worries.

    Thank you for your time.
    Like 0    Dislike 0   3 weeks ago
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