I will admit, I am neither too smart nor mathematically inclined; I hope to be in the near future though. Please assist me with my assignment. Of course you do not necessarily have to answer it. Insight and perspective in tackling this question would be highly appreciated. Thank you!

That's an easy one. First, analyze the 300N force into cartesian axes x and y. The y one is 300*sin25 while the x one is 300*cos25. The negative y part will be removed from the 250N force that is above. From the given picture, we assume that the angle of F from the x axis is also 25 degrees. You then analyze that force too into F*cos25 and F*sin25 for each axis. You add these forces to the respective combined forces from before and gain a new result for each one, with the only unknown force being F. Finally, you use Pythagoras' theorem to calculate the full combined force of the two new combined axes-forces and that big combined force should be 500. Solve by F and you shall get the result. EDIT: According to my calculations the correct answer is F=-244.95N If the angle of F is not 25, you will have two unknown quantities (the angle and F) which means you'll need another equation that connects F and the angle. Since there's no real indication here, the most probable thing is that the angle is indeed 25.

^The problem asks the magnitude and direction of the smallest force F. This angle is unknown. Thus, to solve the problem, you only have to do the vectorial addition of the two known forces and complement the rest with the third one at the same angle of the combined two forces. If I take the horizontal line as 0°, then 250N force has an angle of 90° and the 300N has -25°. The sum of these forces is 298.51N at an angle of 24.379°. It suffices to add the third force of 201.49N at the same angle. You can use a calculator in complex mode to calculate the vectorial addition easily, since complex number and vector use the same principle. I used this instead because it is faster than searching for my graphical calculator: http://www.mathsisfun.com/algebra/vector-calculator.html. You can add the numbers I wrote up there into the blank fields to visualize the vectors. If you need to write out the whole calculating process, use some trigonometry. Like complex number, you can also divide each force into their coordinates and calculate the sum via the sum of the coordinates. Sorry for my bad explanation though, it has been a while since I last used English in Maths.

Damn, you're right :ninja: Well, the first 2 steps still apply. As I've said, you'll just need to figure an additional equation to connect the angle with F and replace it on your combined-force equation.

That's counter-productive though. If he wants to learn how to solve a problem like that on his own, he should be able to go through this procedure. An even better practice is to completely ignore the numbers and replace variables with letters. Solve the problem with pure algebra and then replace all the known ones on your final product to get a result.

Well, the point of studying is to understand the method tho. So yes, replacing those numbers by variables is the best. But I'd use a calculator all days tho. Though it is not like I'm heavily dependent of a calculator; I can do Maths in my head whenever I can.

Hi guys, here's another question that got me phased a bit. Please guide me on how to tackle this question. Question 4.126 https://books.google.com.ph/books?i...d=0CCAQ6AEwAQ#v=onepage&q=pytel 4.126&f=false

Hmm... I don't know if this is usual lever problem or not... (Drawing is not to scale) The equilibrium is reached like this right? With O the center of the surface. I just deduce this because the bar is rotating around O, hence the equilibrium is when vector P passes through O. Other way to deduce is that the two reaction forces applied by the surface on the bar always head toward O (because the surface is cylindrical), and the sum of these two forces have to balance with the weight of the bar in order to reach equilibrium; therefore, the force P and the center O have to be on the same line. If this is true then the problem is pretty simple, it suffices to calculate the length of OG. First of all, calculate the angle OBG via cosine formula (you know the length of OA, OB and AB). With the same formula you can calculate OG. Since OG is vertical, and OAG is an isosceles triangle, the calculation of alpha is easy. The main difficulty of the problem is to know how the equilibrium looks like.

First thing you have to do is understand what the center of mass information is trying to achieve. Your equilibrium here is obvious, but forces on both points of the bar have x and y vectors. Since the center of mass is set like this because of disproportionate density, each part of the two will be subject to the same force: the overall bar's weight divided by 2. The vectors for point A: Fx=m*g*cosθ/2 and Fy=m*g*sinθ/2 while for point B: Fx=m*g*cosθ/2 and Fy=m*g*sinθ/2 as it is expected for the equilibrium and it is confirmed by the geometry of the two points (I don't know the term in english for the relation between them that makes them equal). However, the cosθ and sinθ quantities (while numerically identical to their respective ones between A and B) are algebraically different for each point. For A: cosθ=100/h (where h the hypotenuse of a taken right-angled triangle) whereas for B: cosθ=75/h' (a new triangle hypotenuse). The two hypotenuses combined make a diameter of the cylindrical surface (both horizontal, going from each side towards G). So h+h'=200 => h=200-h' From our previous equations: 100/h = 75/h' => 100/(200-h') = 75/h' => 100h' = 15000-75h' => h' = 85.71 mm And from this you can solve for θ: cosθ = 75/h' => θ = arccos(0.875) => θ = 28.96 degrees The other way above is also good, but too lengthy and involves too much unnecessary geometry. Simple trigonometry (with a tiny bit of geometry) suffices here.

Your result is double of my result. Check if you have multiplied by 2 or forgot to divide by 2 at some point.

You are right, I misscalculated the overall hypotenuse. The horizontal line that connects the 2 hypotenuses of the triangles I combined isn't exactly a diameter (~180 mm).

If only there are words that could express my deepest gratitude for all your guidance, I don't know how to thank you all enough. I hope you wouldn't mind I burden you with another question that phased me a bit as well. Question 7.9 https://books.google.com.ph/books?i...AWliIGgDg&ved=0CBwQ6AEwAA#v=onepage&q&f=false