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.
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...
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...
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...
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...
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...
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...
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...
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...
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...