Recent content by Jud

-A(0.12^2) = -B So, 140MPa = A + - A(0.12^2)/0.1^2

A = -B/0.12^2 A(0.12^2) = -B

I.e. B = A/0.12^2 So, 140MPa = A + (A/0.12^2)/0.1^2?

I don't understand, sorry. Could you elaborate with the values please.

Homework Statement A steel pipe is 200mm internal diameter and 20mm thick. Calculate the safe internal pressure if the allowable stress is not to exceed 140Mpa. Homework Equations σr = A-B/r^2[/B] and σtheta = A+B/r^2 The Attempt at a Solution Right so, I've got two equations...

Sorry but I need bludgeoning in the head because I get 51.6N again, and I cannot see only one step available. -P = (cos 30 + 0.4 sin 30) - 343.35 (sin 30 - 0.4 cos 30) -P = -51.669N P = 51.669N

P ( cos θ - mg sin θ + μs (sin θ + mg cos θ)) = 0 I'm having major trouble transposing. P(-mg sin θ + μs (sin θ + mg cos θ)) = -cos θ P(μs(sin θ + mg cos θ)) = -cos θ + mg sin θ P(sin θ + mg cos θ) = -cos θ + mg sin θ / μs P (sin θ) = (-cos θ + mg sin θ / μs) - mg cos θ P =...

P cos θ - mg sin θ + μs ( P sin θ + mg cos θ) = 0 Still yields the same answer?

Homework Statement Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 35kg and the coefficient of static friction between the crate and the plane is μs = 0.40 2. The attempt at a solution Fx = P cos 30 - 343.35 sin 30 +...

Ok Thank you for the help, greatly appreciated.

so, F1 (800N) x-pos = 0 , y-pos = 800 F2 (600N) x-pos = 600 cos(30) , y-pos = -600 sin (30) sqrt (519.615)^2 + (500)^2 = 721.1 N is that correct?

I'm sure this problem is pretty trivial to most but apologies in advance, I'm just starting out sorry. Homework Statement I'm not sure how to go about working this out due to the 800N The Attempt at a Solution F1 (800N) x-pos = 800 tan (90)?? , y-pos = ? F2 (600N) x-pos...