Recent content by Ryansf98

Ah apologies, I tend to overuse brackets & at times make mistakes as a result. I'm working from a University past paper so I assumed the marking scheme to be correct no matter what as it's a paper from 2014. So in either circumstances with the parenthesis, there is no way that function could...

Question 16. Definitely yeah.

Homework Statement f(x) = Log(cosh(x-1)), find f'(x). Homework EquationsThe Attempt at a Solution f'(x) = [1/cosh(x) - 1] * d/dx [cosh(x) - 1], => f'(x) = sinh(x) / [cosh(x) - 1] Although, my marking scheme says the answer should instead be; cosh(x) + 1 / sinh(x). Can someone explain where...

I ended up redoing some of it. I calculated R = 284,272N, then used a(centripetal) = v^2/K, => a = 0.35 m/s^2. I then used F(centripetal) = ma(centripetal), = 10,408N. Then I done R - F(centripetal) to calculate lift. I got the answer on the sheet, but is that the correct process? I also...

This is where I was confused, as I was using a incorrectly. I tried to use a = v^2/K, but as you both noted I wasn't using it corretly but I don't know how to use it. So this a would be a(centripetal)? The 2m/s^2 is the overall accleration as far as I'm aware.

Homework Statement "Figure 5 depicts a 30 000 kg aircraft climbing at an angle θ = 15˚ when the thrust T = 180 kN. The aircraft’s speed is 300 km/hr and its acceleration is 2m/s2. If the radius of curvature of the path is 20 km (i.e., θ is decreasing), compute the lift and drag forces on the...

Makes sense. Again thanks for your help!

So generally no matter what, each rope has a tension T throughout the system, even if the rope reaches a pulley and looks to go into 2 separate ropes? Thanks for the help!

In fact, there are 3 ropes above the man and pulley, therefore F(up) = 3T. T = 200N as the man exerts this force down the rope and so the rope exerts this force up on him. That means that F(net of the system) = mg - 3T, => 686.7 - (3*200) = 86.7N downwards. F = ma, a= 86.7/70, => a = 1.24 m/s^2?

One rope attached directly to the man and chair, therefore T up? Also, would 200N be T, due to Newtons 3rd Law? Therefore, Net Force Up = 200N?

I realize I jumped a gun a little there as you only suggested a free body diagram for the man and chair. For this, I labelled mg down and T up towards the smaller pulley. Would I in fact insert 200 N here also?

I done that, labelling F(down) as 200N + mg, => F(down total) = 887N. Regarding the tension, I labelled T moving up from the man to the smaller pulley. Would that then split into 1/2T either way up the pulley, or would it remain T? I also labelled T pointing up on the rope which the man pulls...

Thank you by the way! I've used it for sometime looking at existing threads, but I couldn't find an existing problem like this one so thought it'd be a good time to start my own account!

I just read that over and now realize it doesn't make sense, as equal tension & weight would mean an acceleration of 0m/s^2 wouldn't it? What I meant was that tension must be equal throughout the rope.

Homework Statement The question has been screenshotted, along with the given diagram. Basic stuff I know, but I can't seem to get my head around pulley systems. T Homework Equations F(net) = m ⋅ a(system) The Attempt at a Solution I understand the rope to be considered inextensible. Also...