Recent content by cd.riter

Remember: Vy = Voy - gt I think you should revise your Vy => Although for this problem it is not necessary use this Yf = Yo + Voyt - .5gt^2 just think where you want your Yf to be and solve for t

first I made the tensions equal to each other m1*a + Uk*m1*g = m2g - m2 a , then i solved for a a = ((m2 - uk*m1)/(m1+m2)) * ga = (((10.3*.36)-(.336*10.3)/((10.3*.36)+10.3))) * 9.81 = .173

I can't find my error, thanks anyway Doc, I'll just go find my professor, I'm sure he's made an error in programming this, he's done it before

i got .173 m/s/s so i put in the calculator (10.3)(.173) + (.336*10.3*9.81) = Ft

yea i looked over everything many times, I'm beginning to think that its the computer program that's wrong...Well if anything my answer for Ft = 35.73 N, if you are getting anything else please let me know

Here this link really helped me out to solve this http://www.real-world-physics-problems.com/ferris-wheel-physics.html

Block 1 as shown in the link below(url) has a mass of 10.3 kg. The coefficient of static friction between the block and the table is μs = 0.36 and the coefficient of kinetic friction is μk = 0.336. a) What should the mass of the second block be so that m1 is just on the verge of sliding? I...