Recent content by gnits

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    To find the resultant of forces on a lamina

    Thanks to everyone for the help. I too get: $$F_1=8$$ $$F_2=(12-\sqrt{3})$$ $$F_3=(12+\sqrt{3})$$ I will assume here that the answer in the book is wrong. Thanks again, Mitch.
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    To find the resultant of forces on a lamina

    Thanks both for your suggestion of using the original forces for the moment balancing. It is not agreeing with the book at the moment. Would you agree with the following: Taking moments anti-clockwise about D: 8N force along BC has moment: 0 4N force along AB has moment: 4*sqrt(3)a/4 3N...
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    To find the resultant of forces on a lamina

    Could I please ask for help with the following question: A lamina is in the shape of an equilateral triangle ABC, and D, E, F are the midpoints of BC, CA, AB respectively. Forces of magnitude 4N, 8N, 4N, 3N, 3N act along AB, BC, CA, BE, CF respectively, the direction of each force being...
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    To calculate torque on a supported beam

    Thanks very much. That helped me to see it. Moments about Q. Mitch.
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    To calculate torque on a supported beam

    Could I please ask for help with the last part of the following question? I have the first two parts done, answers are: Distance of COG from A = a(1+n)/n and W1 = W(1+n)/3 I can't see how to go about the last part. Here's my diagram for the system prior to the torque L being added: In...
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    How to calculate the magnitudes of three forces around a hexagon

    Thanks jbriggs444 and haruspex for your hints and patience, I see it now and have the right answers. Thanks very much.
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    How to calculate the magnitudes of three forces around a hexagon

    Again, thanks very much for the help\hint, I'm just not seeing through to the solution. I get that if we take the force R which acts along DF and we add to it the clockwise torque 4Pa then the new resultant is equivalent to the force F of the along CA. I know that somehow that should give me an...
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    How to calculate the magnitudes of three forces around a hexagon

    Thanks for the hints, but I still can't see my way to the solution. Given that F is R-plus-an-additionsal-pure-torque and that we a told that F is parallel to R then, in terms of forces, is not F = R? Given any relationship between F and R I am still unsure how I could form a torque equation...
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    How to calculate the magnitudes of three forces around a hexagon

    Can anyone please help me with the following? Three forces which act along the sides AB, BC and CD of a regular hexagon ABCDEF of side 2a, have a resultant which acts along DF. When a couple of 4Pa in the sense CBA is added in the plane of the hexagon, the resultant acts along CA. Find the...
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    To find the magnitude of the resultant of forces around a triangle

    Thanks very much for your reply and confirmation of the answer, Mitch.
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    To find the magnitude of the resultant of forces around a triangle

    But aren't those ratios predicated on the assumption that the resultant is a couple? The phrasing of the second part doesn't suggest to me that I can assume a resultant couple now that the force along AC has been reversed. EDIT: Ok I see that the result about the ratio of P, Q, and R is indeed...
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    To find the magnitude of the resultant of forces around a triangle

    Could I please ask for advice with the following: ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. a) Find the ratios P:Q:R if their resultant is a couple. b) If the force along the directed line...
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    To find the ratio of forces around a right triangle

    Thanks, I see my error. Mitch.
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    To find the ratio of forces around a right triangle

    Could I please ask for help with the following: ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. Find the ratios P:Q:R if their resultant is a couple. Book answer is 4 : 3 : 5 Here's my diagram...
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    Forces proportional to the sides of a quadrilateral result in a couple

    Sorry, yes forgot to mention that the help I received also let me solve this part. Thanks, Mitch.
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