Recent content by kaienx

I saw my mistake! Handling big numbers really are confusing. I have edited the first post for clarity's sake. 1 = 619² cos²β + 60² + 2(619cosβ * 60) + 619² sin²β + 882.9² - 2(619sinβ *882.9))/ 313.9² From there: 1 = 619² (cos²β + sin²β) + 60² + 882.9² + 2(619cosβ * 60) - 2(619sinβ *882.9) /...

my last formula? meaning sin2α+cos2α=1? is it?

I've done some simplifying and adding up based on your guidance and got: 783112.41 + 619²(sin²β + cos²β) + 74280 (cosβ + sinβ) / 313.9².

could you elaborate further on (simplify by replacing the sum of the square terms by 6192) ?

Homework Statement I have 3 masses (Fα, Fβ & Fg) with 2 pulleys, and a wind variable which is in static equilibrium. I have already calculated the appropriate forces for the 3 masses by multiplying it with 9.81m/s² (gravity). Fwind = 60N Fα = 313.9N Fβ = 619N Fg = 882.9N I'm required to find...

The distance between pulleys meaning, the red line? (See picture for reference). If I were to have the values, can I assume that I will be able to solve using basic trigonometric formulas seeing that I have the angles needed? (TOA CAH SOH).

@insightful yes that is correct, it should be alpha, apologies as I myself got confused with the whole question. But thanks for pointing that out! :) Anyway, there is no way to find out the displacement from the first equilibrium position to the second equilibrium position with just those two...

Thank you! Took me a whole day to understand how did you obtain those values and I think I got the accurate answers using different masses on M3. So the new position of the M2 is determined from the new tensions from M3, am I right?

Yes apologies it is supposed to be Fx = 508 * cos (α) - 608 * cos (β) = 0; Fy = (508 * sin (α) + 608 * sin (β)) - 900 = 0 (α) = cos-1(608 cos (β) / 508) = cos-1(1.2 cos (β)) replace (a) to Fy's equation I get: 508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) - 900 = 0; 508 * sin(cos-1(1.2 cos...

So it will be something like, Fx = 608 * cos (α) - 508 * cos (β) = 0; Fy = (608 * sin (α) + 508 * sin (β)) - 900 = 0 I then used algebra to solve for Fx and got (α) = cos-1(508 cos (β) / 608) = cos-1(0.84 cos (β) This part I am stuck as it got a little tricky honestly. does cos-1 cancel out...

Meaning? Which angles? is it the α and β? I think my train of thought is stuck as I really do not know how to find the new α and β. So my calculations was wrong? No wonder!

I can only imagine that the affected mass (M3) will go down, thus M2 will go north east due to the resultant force applied from M3, which then M1 will have to move up to accommodate to the movement of M2. Here's what I got so far: Fx = 608 * cos (62.5) - 508 * cos (62.5) = 234.6 (rounded...

Sorry for not being clear, it is to find a position of the masses after the addition of 10kg at M3 at equilibrium. However I am just curious of the instantaneous acceleration that will occur during the addition of 10kg mass (thus the displacement/velocity/acceleration). That's why I am kind of...

For part 2, the necessary condition for an equilibrium on the system is that all net forces must equal to 0, however, when I calculated after adding 10kg on M3, wouldn't the Fx and Fy for M2 also changes (correct me if I'm wrong), which means M2 will travel along the rope of M3, which will pull...

Good day everyone, I found this forum while on a quest to complete an application that requires physics (from scratch). Am a developer at a local polytechnic. Pleased to meet everyone! Anyone from SEA here?