Recent content by lemonpie

okay... so what am i doing wrong?

i understand that it's a vector, but where do i set the origin? if i set it at m4, i still don't get the answer that i want. i get M = 2m, which is still not correct.

oh, for some reason i thought it said m = m4. Fnet = 0 = F1,4 + F2,4 + F3,4 0 = GMm4/r2 + Gmm4/r2 + Gmm4/r2 0 = M + 2m (canceling out G, m4, and r2) M = -2m um...

Homework Statement Two spheres of mass m and a third sphere of mass M form an equilateral triangle, and a fourth sphere of mass m4 is at the center of the triangle. The net gravitational force on that central sphere from the three other spheres is zero. (a) What is M in terms of m? (b) If we...

okay, i think i found the right answer, but I'm not really sure why it's right. net torque = (75)(9.8)3sin30 + (10)(9.8)2.5sin30 - 4.33F3 = 0 1102.5 + 122.5 - 4.33F3 = 0 F3 = 282.9 N the part i don't understand is (4.33)(F3). why does r = 4.33? that would mean that r just goes straight down...

1. i chose the base of the ladder as the origin, i guess because i don't know that force. 2. so i think this is what i don't understand. my forces basically go straight downward, vertically. so the direction of r has to be... horizontal? 3. so i can only consider the forces acting on... the...

Homework Statement A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long. He places one end on the ground 2.5 m from a a wall, rests the upper end against a cracked window and climbs the ladder. He is 3m up along the ladder when the window breaks. Neglect friction between the ladder and...

um... I = 1/12ML^2 + M(1/2L)^2 I = 1/12ML^2 + 1/4ML^2 I = 1/12ML^2 + 3/12ML^2 I = 4/12ML^2 I = 1/3ML^2 fortunately (or unfortunately), there weren't any questions about the parallel axis theorem on the test. i got a B+, not as good as i'd hoped, but considering how little i knew, maybe i...

okay, but this problem: https://www.physicsforums.com/showthread.php?t=116823 rotation is about point A, and it doesn't require parallel axis theorem. sorry! I'm trying to understand!

Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12 d^2 + M(1/2 d)^2) + md^2 + (M/12 d^2 + M(3d/2)^2) + m(2d)^2

so... d/2 and 3d/2 are for the Mh^2 not Icom? that seems weird to me. wait let me try.

actually, why do you have to use the parallel axis theorem here? why can't you just add up the rotational inertias for rods and particles? someone please help me out soon! thanks!

sorry, i had class. okay, how about this? Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12(1/2)d^2 + Md^2) + md^2 + (M/12(3d/2)^2 + Md^2) + m(2d)^2

sigh. i guess i would like to treat them as one rod. so this is my answer (i noticed that i wrote a wrong number, so I'm fixing it here): Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2 i think the particles are okay, but not the rods. is...

oh. h means distance from O to com. not just d. Inet = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2 ?