Recent content by Seaborgium

Homework Statement By considering A x (B x A) resolve vector B into a component parallel to a given vector A and a component perpendicular to a given vector A. Homework Equations a x (b x c) = b (a ⋅ c) - c (a ⋅ b) The Attempt at a Solution I've applied the triple product expansion and...

Wowee. That felt like it took far longer than it should have. Thanks again for all the help!

Oh damn, overdosing on triangles tonight. It's R/(tan(θ/2)) from C?

Distance of R from C, perpendicular to the stick - so R.mg/2 sin(θ/2) for that component. The whole line for torque goes 3R.T.cos(θ/2) - 2R.mg/2 .sin(θ/2) - R.mg/2 .sin(θ/2) = 0, components are Tension, ground-contact and cylinder contact respectively

Oops, I went off the assumption that the horizontal components of the contact forces on the cylinder would cancel each other out and neglected to include them in the torque balance. This time I got T = mg/2 tan(θ/2). Does this look more correct?

I ended up with T = mg/3 tan(θ/2) , which seems to make sense. Thanks to both of you for all the help, I feel like I understand this topic much better now!

Oh Jesus, I see now. I've never seen a problem of this nature, so I'd assumed it would be like the others, without realising the cylinder wasn't simply experiencing one or two forces. The three forces acting upon it would be gravity, then a normal contact force from each of the sticks? Now I get...

They're equal and opposite? Disappointed it took me that long to realize that. So the normal contact force is equal to its horizontal and vertical component vectors added together - neither of which I feel like I understand how to find, even having done simpler static systems before. It's...

mg/2 would be larger, seeing as the contact force is at an angle to the vertical that the mg/2 vector points along? Sorry if I'm missing obvious stuff, I'm still a bit new to Physics.

Thinking it's mg/2. Would the contact force be (mg/2)cos(θ/2)?

I've got it! I need to resolve moments around the pivot for both sticks, which should be equal to zero as neither is rotating? Makes a huge amount more sense now. Still no idea where to go on defining the normal contact force between the cylinder and the sticks, though.

I haven't drawn the thingy to scale - the top of the cylinder is above the string. Fixed the error with mg/2. I thought it looked fishy when noting it on. Also, realized its mgcos(Φ) (for Φ = θ/2) and not sin. The ground forces completely elude me though. There isn't any friction involved, as...

Hey, thanks for the reply! I drew the diagram and noted as many forces as I could identify on the diagram - tensions, normals and gravity. I then drew a vector triangle out with the three vectors on, as Tension and Gravity are orthogonal to each other in this case, giving me a triangle with the...

Homework Statement "A cylinder of mass m and radius R is lodged between crossed sticks that make an angle θ with each other. The crossed sticks, each of negligible mass, are connected at the point C, with AC=BC=2R and CD=CE=3R. Determine the tension in the string at DE. Assume the floor is...