Recent content by dgibbs

Homework Statement Spiderman, whose mass is 80.0 kg, is dangling on the free end of a 12.0 m long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a...

This makes a lot of sense, and living in NYC I can definitely relate to what your saying with the elevators. So my answer is going to be the Normal Force... Fn = gravity + acceleration = (-mg) + (v^2/r) **My next question is this...in this equation gravity will be negative because it is...

So his weight will be the acceleration downwards? v = velocity? r = radius ?

I believe my teacher is talking about using units such as lbs and feet.

1. Homework Statement The Pilot of an airplane executes a loop-the-loop manuever in a vertical circle. The airplane's speed is 300 mi/h at the top of the loop and 450 mi/h at the bottom. The radius of the circle is 1200 ft. Use American Engineering units in this problem. A) The pilot's...

any help please?

Is the right equation... Fn = -mg + mv^2/r ??

Homework Statement The Pilot of an airplane executes a loop-the-loop manuever in a vertical circle. The airplane's speed is 300 mi/h at the top of the loop and 450 mi/h at the bottom. The radius of the circle is 1200 ft. Use American Engineering units in this problem. A) The pilot's...

Okay. So we have accounted for all the Forces. Ff=μN = (0.250)(Pcos50.0) Fp = - Psin50.0 Fg = - mg(m) = -29.4N Now I can isolate and solve for P...thank you very much Ms. Sandy Bridge, you were extremely helpful! Have a great evening

Nevermind I think I understand the Fn part, Fn is the positive force.

Okay so to sum up: Fmg = (-) because it is going downwards Fpsin50.0 = (-) because it is going downwards Fn = Fpcos50.0 **but is not (-)? or is also (-)** So if these are the forces, I am still confused on one more topic...where does the friction coefficient come into play if Fn is a function...

Okay, so the force P must be negative to show its relationship with gravity and I guess I should also add Fpcos50.0 to the ƩF to represent the normal force?

The force P is acting horizontally at an angle of 50.0 degrees. The Force P seems to be merely holding the mass in place neither pushing it up or helping it slide down. It is possible that box could be sliding up or down depending on the magnitude of P.

Okay, so is it... Fμ(Fpsin(50.0)) + Fpsin(50.0) - mg = 0 (0.250)(Psin(50.0) + Psin(50.0) - (29.4N) = 0 Then solve for P?

The sum of the forces in the Y direction is 0 right... So ƩFy = 0 F(static friction coefficient) + Fpsinθ + -mg = 0