Recent content by tomul

AB will be twice OM, and M will move up and to the right If AB is the radius of the circumcircle of the triangle, ##\frac{L}{2}## then ##OM=\frac{L}{4}##. This prompted me to split the height into two sections being sides of two triangles, one with hypotenuse MB and another OM and I've drawn a...

Oh I see. I hadn't seen differentiation used in the context of a virtual work problem before. This is the approach I've taken as suggested by haruspex but I don't know how to proceed with it. As I say, geometry is a really weak point for me.

Thank you, I understand your argument now. However, I still can't get the correct expression for ##H(\alpha)##. Geometry is a real weak point for me. The problem also states I should use the principle of virtual work, so I don't think that method would be sufficient.

Apologies for the crudely drawn sketch. Is this along the right lines? ##h(\alpha) = \frac{L}{2} sin(\alpha)##

The circle is centred on the centre of the hypotenuse and passes through the lower vertex of the triangle where the planes intersect. So doesn't that meant that the height about this point is simply the radius of the circle, half the length of the bar? I've misunderstood something because the...

Okay, applying the sine rule: ##\frac{sin\alpha}{h} = \frac{sin\beta}{l}## ##h = \frac{l sin\alpha}{sin\beta}## Since ##\alpha## < 90 deg, ##sin\alpha## increases with ##\alpha##, and since ##sine\alpha## is proportional to ##h##, ##h## will increase. The only problem with this is if ##l## or...

It increases. Okay so I need to just consider this central displacement and the weight acting through it? Well I'm having trouble with that too which is in my reply to ergospherical... Okay, I've applied the sine rule to find ##h(\alpha)## ##\frac{sin\alpha}{h(\alpha)} =...

I imagined the bar slipping along a virtual displacement, with the top and the bottom slipping by equal amounts. Since the planes are orthogonal, I took the components of these displacements: the lower end is displaced Lsinα in the direction of the right plane and by Lcosα in the direction of...

My attempt was to calculate the acceleration of M2 as the acceleration of M2 if it were the only mass in the system, minus the component of M1's acceleration along the slope. And then I would divide the whole thing by 2 to get the acceleration for just one of the two masses@ a = 1/2 ( g -...