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In order for the man to move at constant velocity the net force on him must be zero. He can't move at a constant velocity if there is an unbalanced force on the right (think about it: the right side of the rope would accelerate but the left side wouldn't? would the rope not rip in this case?)...

Homework Statement I don't understand why when we derive the velocity equation of motion in polar coordinates we start with position equal to R times R hat and not (theta times theta hat + R times R hat). Homework Equations none really.. The Attempt at a Solution Is there an assumption I'm...

Hey Guys, I'm currently finishing up my B.S in Physics (going into the last semester of Junior year) and plan on getting a Masters in Computer Science concentrating on software engineering. I'm curious if anyone else has made a similar career choice and how it worked out. Also, which...

kinematic equation = equation of motion (this usage should be in your book)

My apologizes, llello is right.

Look for a kinematic equation for position not velocity.

a book is dropped from the ground?

You're not "adding vectors." You need to apply conservation of momentum. Momentum is a vector quantity. Thus if you know the momentum vector before the collision, the sum of the individual momenta afterwards must equal what?

I don't believe law of cosines will get you anywhere. Think about breaking your velocity vectors into their components.

To have a visual representation of light, light would have to reflect off light and hit your eye. This doesn't occur in reality as far as I'm aware.

Did you put your angle in radiants?

You must attempt to make a solution before receiving assistance.

The purpose of the third rope is to most likely stop the ring from accelerating correct?

Normal to the wall goes straight through the axis of rotation. right?

150 degrees is the angle between the ladder and the wall (outside the triangle, inside the triangle it is 30). What is the angle between the force normal to the wall and the lever arm?

Before we continue let's explain why the normal force on the ladder by the wall cancels. What is the angle between the force vector and the lever arm? (lever arm is just another term for "distance from axis of rotation to the force.")

Where it touches the wall will help the most because it will save us from having to calculate the force the wall exerts on the ladder. So, which forces are acting on the ladder? And where along the ladder are these forces being applied? (What are their lever arm values?)

Correct, it wouldn't produce a torque because the angle between the lever arm and the friction force would be zero. T = L\times F = |F||L|\sin{\theta }=0 What is a more ideal point to place our axis? One in which the friction force does produce a torque.

If we decided to take the bottom point of the ladder as our axis of rotation would the friction force produce any torque?

We know that: \Sigma T = 0 \Sigma F = 0 What is the best choice for the axis of rotation?

you know that: \frac{sin(53)}{H_a} - with H_a being the height after is equal to what?

I must dispute berkeman's method. "A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?" He is pushing the wheel barrow in the horizontal direction (not up an incline). the vertical displacement is zero...

Total work done = 0 the man does positive work and gravity does negative work of equal magnitude. Thus resulting in zero net work on the box.

I'm saying that: Ke_f - Ke_i = 0 = \Sigma W because the box starts at rest and finished at rest the total work done on the box better be zero.

I'm not saying the answer is zero. We are being asked for the work done by the man; I'm saying the total work done is zero. \Sigma W=fd=\Delta Ke=0 -(at the top of the stairs) Now which forces do work on the box?

Let's start with problem #1: Would you not agree that because the box has no velocity at the top of the stairs the total work done on the box is zero? Also, which forces do work on the box? When can work be negative? Only when the force is opposite the displacement.

Your problem is stated very poorly. You assert that the cart has only two wheels but where are the two wheels located? One wheel in front of the other? Both side by side at the rear? Both side by side in the front? Both side by side below the center of mass? regardless, remember that because...

you found the total kinetic energy at the bottom. Because the sphere is rolling what is the total kinetic energy equal to? also note: \omega=\frac{v}{r}

mass was something like .00022 & gravity was .0088

I read that book and liked it a lot (although i didn't understand most of it), Thanks!