Search results

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...

Thanks very much. That helped me to see it. Moments about Q. Mitch.

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...

Thanks jbriggs444 and haruspex for your hints and patience, I see it now and have the right answers. Thanks very much.

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...

Thanks very much for your reply and confirmation of the answer, Mitch.

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...

Thanks, I see my error. Mitch.

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...

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

Thank you both for your help. You enabled me to see how to understand and solve the question. It really puzzled me at first. Indeed it is essential that the magnitudes of the forces be proportional to the lengths of the sides so that one can legitimately form the force polygon. That was the...

Can I please ask for help with the following: Forces proportional to the sides of a quadrilateral taken in order act respectively along those sides. Prove that the resultant of the system is a couple whose magnitude is represented by twice the area of the quadrilateral. Not sure where to start...

And I misread the book answer as 22/21 when it is 92/21

Thanks so much (yes it was a typo in posting). I misread the question as saying that the forces were replaced by F4 which was to be equilvalent to the initial forces, but of course that's not what it is asking, it is saying that F4 is added in to the system to create equilibrium. Thanks again...

We know from the first part that: F1 = 8i + 6j F2 = 6i-3j F3 = 7i-24j Resolving horizontally and vertically gives us F4 = -21i+21j Taking anti-clockwise moments about (-2,4) gives: 8*6 + 6*3 + 7*4 - 21 * 6 = 21 * (y' - 4) + 21 * 2 This leads to y' = -8/21 and so to an equation of y = -x -...

Could any one please help with the following? Three forces of magnitudes 10, 3 * sqrt(5) and 25 newtons act along lines whose vector equations are respectively: r1 = i - 2j + k(4i + 3j) , r2 = -2i + 4j + k(2i - j) and r3 = 4i + k(7i - 24j) A fourth force F4 is introduced which reduces the...

Just finished writing and posting this, then after all that effort, I found my mistake - an algebraic slip. The method works.

Could I please ask for any help with the following question: Here's my attempt: (i and j are unit vectors in the directions of east and north respectively) (apologies that LaTeX is simply not working for me, I'll label the angles in each case T and P as shown in my diagram) Let the...

Thanks for your reply, it helped me see my error. I was indeed wrong to say that the max value will occur when cos(wt) = 0 and sin(wt) = 1 As I have the sum of three terms and the first two are positive, I will need to maximize the last term. I wrongly stated this max as 1 but it is in fact...

Could I please ask for help regarding the final part of the following question: It is the very last part, to find v in terms of u. So I have that the velocity of the midpoint of XY is: V_m = (u/2) i + (u/2) j I let the position vector of P be: r_p = cos(wt) i + sin(wt) j (w = angular...

Thanks all for your help. I see my error now. Thanks very much for the quick help.

Could I please ask for help on the last part of this question: So, part b, I get the right time but not the right distance. Book answers are: distance = 1/6 and time = a/V. Here's my (faulty?) reasoning (LaTeX isn't working for me): The boat is steered due east and so would have a velocity...

Thanks very much. That was indeed my mistake. I made it each time I tried to solve the problem. Thanks again, Mitch.

Can anyone please help me see if my reasoning is correct regarding the following question? I'll just solve for the case where the dinghy tracks so as to just 'touch' the exclusion zone on the 'high' side So, in the diagram below: The dinghy tracks along the red path, inclined at x degrees...

Thanks very much for your reply, I appreciate it.