Recent content by sliced cheese

As I understand it, the reduction in slope of the stress-strain curve is, in fact due to a small area bearing the stress. Per unit area, even after the yield, it does take more force to induce the same amount of strain. However, due to the area the force is applied to being smaller, the...

I very much agree.

Ok, after a failed attempt at long division, a failed attempt at synthetic division, a realization of my mistake, and then a successful attempt at synthetic division I end up with (x-a)(x^2+ax-2a^2)=0. So using the quadratic formula on my second factor I got the roots x=a and x=-2a. So, they...

y-a^3=3a^2(x-a) y-a^3=3a^2(x)-3a^3 y=3a^2(x)-3a^3+a^3 y=3a^2x-2a^3 set equal to y=x^3 x^3=3a^2x-2a^3 x^3=a^2(3x-2a) new step, probably no more useful... x^3=a^3(3x/a-2)

So I rearranged and set those two equations equal to one another, I end up with x^3=a^2(3x-2a), not sure where to go from there, I don't see any difference of cubes to factor like you mentioned earlier.

In my equation, y is on the left side y=a^3 (for the point (a,a^3)) I was confused by the formatting of your equation, I don't think I've ever actually had to use it before. I see the connection now between it and the one I have used to find slope between 2 points (m=(y2-y1)/(x2-x1)). I'll go...

Ok, wasn't sure because I've been yelled at on forums for not using the search function before making a thread. Say I use the point P(a,a^3) (or x0's or whatever), I don't understand why the equation is (y-a^3)=(3a^2)(x-a). I would substitute the point into the equation for a line y=mx+b and...

The tangent at a point P on the curve y = x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.y=x^3, dy/dx=3x^2, y=mx+bI tried setting the derivative equal to the original function to see when they intersect and go from...

Ahh I see. Thanks for all your help :)

Ok, so Force of friction = coefficient of friction * mass * 9.81 I know mass, but not the coefficient of friction. How can Delta E = 0 if it starts with some kinetic energy and ends with no energy? Kinetic energy at the moment of release is 0.5 * 20 * 0.885^2 = 7.83 N*m which is a watt? I think...

Hmm, isn't work force multiplied by distance? But I don't know the force, or the work done.

Care to explain a little more please? I can find the initial kinetic energy... can I get from there to the force of friction knowing the distance it took to stop?

1. A curling stone with a mass 20.0kg leaves the curler's hand at a speed of 0.885m/s. It slides 31.5m down the rink before coming to rest. Find the average force of friction acting on the stone. 2. So far I have (I think) net force, acceleration, average velocity and time. I need...