Recent content by mistermill

Gotcha. Fapplied at the bottom goes toward the wall, causing a torque in the same direction as the torque from the person. Now we have two unknowns! Because we don't know F normal. A good reason to move the pivot point.

Well. I drew my torque from the person straight down. I drew the applied horizontal force at the bottom as straight out from the wall. So those two torques do act opposite one another. When I also include a normal force at the bottom of the ladder, it has a torque opposite the torque from...

Is it the Normal force acting perpendicular to the floor?

So there is a torque missing when the pivot point is at the top, which accounts for the different answers.

Right, so if the contact with the wall is frictionless, does that mean it can't be used as the pivot point?

Not according to the answer keys I have come across. But what about the normal forces at the top and bottom of the ladder? they also don't show up in any of the answer keys. Perhaps that is the 'applied force' at the bottom?

A weightless ladder 7.0 m long rests against a frictionless wall at an angle of 65° ablove the horizontal. A 72 kg person is 1.2m from the top of the lass. What horizontal force at the bottom of the ladder is required to keep it from slipping? Στ=0 τ=FsinθdSolutions for this question always...

Homework Statement My questions reads: An electron is pulled away from a fixed charge of 1.3μC. The electron is moved from the positive charge to 4.0 cm away from the charge. If the electron is released from the 4.0 c mark, what is the max velocity of the electron? Homework...

Homework Statement 11. Two parallel conductors, 1.0 cm apart, each carry 10 A of current in opposite directions. What is the magnetic field strength at the midpoint between these wires?Homework Equations B = μI/ 2πrThe Attempt at a Solution I know that the magnetic field is doubled because...

Homework Statement Find the electric field midway between a -3.0nC charge and a -5.0 nC charge that are separated by 0.60 m. Homework Equations ε= kq/r2 The Attempt at a Solution net ε = ε1 + 2 I drew my -3.0 charge on the left and my -5.0 charge on the right midway in...

I like it when things are over my head. I appreciate the insight. I have read ahead to tangential speeds. And I understand there is a difference in a vertical circle with a rigid object and a vertical circle with a string. I just find it weird that my text points the Tension down at the...

Homework Statement In our lab, we times a cart going 1m. The time was 9.49 s. The cart was pulled by a string, over a pulley, attached to a free-falling mass. The purpose was to calculate the μ, the coefficient of friction. Homework Equations d = vit + 1/2at^{2} a = (vf-vi)/t...

Does the string exert a 'normal' force? Why do we not account for it?

When you said 'away' I got startled and thought you meant away from the centre. Whew! So the FBD are right. And if I use Fc = Fg + Ft for the top and rearrange Ft = Fc - Fg And put in the signs -Ft = -Fc - -Fg and it comes out negative, then the question is physically...

So the intro physics that we are taught is basically impossible. Does everyone agree? All you people that are smarter than me, please chime in!