Recent content by tissuejkl

Is there a reason why the derivative in the second term should be the partial derivative ## \frac{\partial L}{\partial v^2} ##, then?

Yes. Sorry, I should have made that clearer.

Homework Statement ## L (v^2 + 2 \pmb{v} \cdot \pmb{ \epsilon } ~ + \pmb{ \epsilon} ^2)##, where ## \pmb{\epsilon}## is infinitesimal and ##\pmb{v}## is a constant vector (## v^2 ## here means ## \pmb{v} \cdot \pmb{v} ## ), must be expanded in terms of powers of ## \pmb{\epsilon} ## to give...

In the line Rsinβ<=μ(Rcos(β)+Fsin(β))+Fcos(β)<=μR(cos(β)+sin(β))+Fcos(β), I used the fact that F/R<=1 (which implies F<=R) as follows: Rsinβ<=μ(Rcos(β)+Fsin(β))+Fcos(β)=μRcos(β)+μFsin(β)+Fcos(β)<=μRcos(β)+μRsin(β)+Fcos(β)=μR(cos(β)+sin(β))+Fcos(β) Yes, I agree that any friction between the...

Okay, I think I have gotten a solution along the lines of your hint, though I don't believe we need to neglect friction. Considering the forces on the wedge: we have the normal contact force from the cylinder, R; the force of friction, F (parallel to the wedge face and acting in a direction...

Yes, I have drawn a free-body diagram. There are 5 forces: the weight of the cylinder; the normal contact force and parallel friction force from where the wedge touches the cylinder; the normal contact force and parallel friction force from where the cylinder touches the slope. Taking moments...

Yes, β must be at least α so that the wedge face is above the horizontal. With regards to the upper bound on β, something of the form tan(β)<μ must be shown, but I still can't find a way of eliminating the various forces in the ratio sinβ/cosβ

Homework Statement A heavy uniform cylindrical drum is placed, with its axis horizontal, on a slope inclined at an angle α to the horizontal. It is prevented from sliding or rolling down the slope by a triangular wedge. The weight of the wedge is negligible compared with the weight of the drum...