Recent content by kyiydnlm

you need to show us top end of the box.

2VdV/dt = dV^2/dt

r_normal is unit normal vector of your flat square. F_support is force on the ball by the flat square. F_equavilent is equavilent force applied on the ball. M_r is rotational matrix. theta_x, theta_y, theta_z are rotation angle about x, y, and z axis respectively. look, support force on...

OK, you can think about it this way. If you put a box on a smooth surface and a same box on a rough surface. and you push them with same force, which one moves faster? or you try to keep them with same speed, which one needs larger force? now, put two boxes on same surface, but one surface is...

draw free-body diagrams first; then you will find the answer by yourself.

try perspective projection.

hi, chadlee88, try vector if you do not want too much trouble, especially when you are programming. \left[\begin{array}{c}M_{r}\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\0&cos(\theta_{x})&-sin(\theta_{x})\\0&sin(\theta_{x})&cos(\theta_{x})\end{array}\right]...

\frac{L^2}{4} \theta K+J\ddot{\theta} = 0 J = \frac{1}{12}ML^2 case 1: K = K1 case 2: K = K1+K2 Assuming uniform seesaw, pivot is center of mass (also geometry center). If otherwise seesaw is not uniform, J will change accordingly. If pivot is not geometry center, the equation will be...

If you replace g by \frac{GM}{r^2}, things become much easier.

of course, you can always use its definition: I = \int_{M} r^2dm

L_{B} = 0.2m (distance between center of mass and axis if you use moment of inertia on center) I = \frac{1}{12}ML^2 + M(0.2)^2

yeah, you can say that.

On the pic you provided, ignore the bottom element and let's say index 1 to 4 from left to right. 1 and 3 are on exactly same plane; they are same thing. 2 and 4 on same plane, they are identical. The stresses of each of the pairs are in same direction. 2 and 4 are on principle plane; maximum...

hi, staetualex, take a little time on chain rules. It's not dx^2/dv; it's vdv/dx.

Since you can find 2\varphi, the angle between principle stress and normal stress is \varphi. If you draw your mohr's cicle, the rest will be only a metter of rotating \varphi degress from horizontal axis to find principle stress direction. In geometry, round angle is half of centre angle...