Recent content by dwangus

The contact point of the object and the cliff?

Homework Statement There are two objects rolling down a hill of incline theta, one is a sphere and one is a disk, each of equal radius and mass. Which one gets down first and how much faster than the other? What's the coefficient of static friction of the hill? Homework Equations Moment...

^^^ welp then, looks like a bunch of my other problems are wrong then

Like I said, can anyone just tell me if I'm right or wrong? It's ok if you don't tell me how I'm wrong, I want it that way. This way, I can chance upon the answer myself.

...how am I making this difficult?

Problem 7: (a-c: force is not counted in problem yet) a) Since it has velocity up the incline, the frictional force is added (helping) to mgsin30 30cos(30)(0.3) + 30sin(30) = 3a a = 7.598 b) Velocity is down the incline, so frictional force is subtracted (opposing) from mgsin30 a = 2.402...

Problem 6: Acceleration of mass A = Acceleration of mass B (downwards is considered negative in the case of mass B for convenience) 20 - T = 2a(B) T - μ(k)m(A)g = 2a(A) T - 4 = 2a(A) ----------------- 16 = 4a a(A and B) = 4m/s^s T = 12N (mass of C and A)g + [(mass of B)(g) - T] =...

Problem 5: Gravity Force exerted from 2kg mass and 20kg mass = 220N Total Normal Force = Gravity Force + Net Force of 8kg mass = 220 + 40 = 260N I feel as though my reasoning is far too simple to have gotten this one right.

Problem 4: FBD of mass m = -Force of mg downwards -Force of μ(s)F(app) upwards to cancel out mg -Force applied to the right -Opposite normal force (N1) exerted by mass M to the left μ(s)F(app) = 50 F(app) = 125 F(app) - N1 = 5a(m) N1 = 10a(M) F(app) = 15a Acceleration of pair =...

Problem 3 (Previous Problems were 1 and 2): Acceleration (A) = -Acceleration (B) T - m(A)g = 5a(A) T - m(B)g = 4a(B) T - 50 = 5a(A) -----> 50 - T = -5a(A) = 5a(B) T - 40 = 4a(B) --------------------------- 10 = 9a(B) Acceleration of B = 10/9 Acceleration of A = -10/9

For the 6 kg mass, T + N (exerted from the 3kg mass) = 60N For the 3 kg mass, T = 30 + N (exerted from the 6kg mass) 2T = 90 T (also from the pin) = 45N Normal Force = 15N

Tension in second rope between 3kg and 4kg abbreviated as = T2 40 - T2 = 4a T2 - 30 = 3a a = 10/7 T2 = 240/7 T = T2 + mg T = 240/7 + 40 = 520/7

Ok, but... In order to complete physics hw, being spoon-fed answers just doesn't cut it. Amongst all subjects, physics requires work shown AND an answer to even remotely do well in class. So if you're concerned that I'll just take your answers and run with them and put them on my worksheet and...

^^^ I don't know if I'm stuck, that's why I need to know whether or not my answers are wrong. I self-teach myself a lot, let's just put it that way.

Add. Problem