Recent content by JJB11

Ah, that's a good idea. You then get the gradient of the tangent between the centre of gravity and the pivot point and because m=tan theta then tan theta = (9a/2b). I've also got the answer to part (b). If DT is vertical then it is parallel to the weight which lies along line y=b/2. Therefore...

What do the points of B and T end up being? I've got T as the old centre of gravity (0,b/2) but I'm not sure about B at all...

Sorry, that was silly of me: A and D are actually (-2a,0) and (-a,b).

A uniform lamina ABCD is in the shape of a parallelogram and its mass per unit area is u. Axes Ox, Oy are chosen, with O the point on AB such that AO = 2/3AB. Points A, B, C and D have coordinates (-2a, O), (a, O), (2a, b) and (-a, b) respectively. Find the x and y coordinates of the centre...

3.(a) A particle P, of mass 0.03kg, is attached to one end of a light inextensible string OP, of length 1 m. The other end of the string is attached to a fixed point O. The particle moves in a horizontal circle, with centre vertically below O, at an angular speed of 2 revolutions per second. The...