Recent content by hyperddude

Oh, the force I found, \frac{10}{\pi} wouldn't be the tension because the force gets divided evenly into 2. Then the tension would be \frac{5}{\pi}?

Homework Statement Let's say there's a uniform circular wire with radius r. There is a uniform force pushing outward from the center in all directions of the wire. This total force is 10N. What is the tension in the wire? Homework Equations N/A The Attempt at a Solution Let's say the...

Thanks!

Oh, good catch. I believe the mass should be 1.2*vwh\sin{45°}. So the force is 1.2v^2wh?

Homework Statement Say there is a wall with width w meters and height h meters. There is a wind with velocity v m/s blowing on the wall at a 45 degree angle. The collisions of the air molecules with the wall are perfectly elastic. What is the magnitude of the force on the wall? (The density...

Yes, but I'm interested in finding what the horizontal force for one of the supports is.

To anyone who saw my previous thread, yes, this is quite similar to it :-p. Homework Statement Given a rectangle, say a painting, with with mass m, height h, and width w with two point supports to a wall at its two upper corners, what force does each support exert? Homework...

Oh, good catch. No wonder the rest of my problem wasn't working out. Thanks!

Alright. The torque due to the component of gravity perpendicular to the radius is mgL\sin{θ}. Since \sin{θ}=\frac{w}{\sqrt{w^2+h^2}} and L=\frac{\sqrt{w^2+h^2}}{2}, the torque is therefore \frac{mgw}{2}. This torque is equal to Iα and I=\frac{m(w^2+h^2)}{3}. So, α=\frac{3gw}{2(w^2+h^2)}...

By radial, you mean in the direction of the radius, right? Is that the only thing I need to consider in finding the force by the pivot?

My question was what the force from the pivot was. You mentioned in a previous post that there would be a component acting in the radial direction (thus making the net force in the radial direction 0). What about perpendicular to the radius? Does that mean the component of the pivot's force...

So does the linear acceleration (from the pivot, I assume) contribute to the angular acceleration from gravity? I don't quite see how they are related.

Woops, θ=\tan^{-1}(w/h), not h/w. With the moment of inertia, I can figure out the motion of this rectangle. How does that relate to the force that the pivot supplies? Does it supply a force equal to the component of the force of gravity on the rectangle, that is perpendicular to the component...

To do this, I'm thinking that the "lever arm" is the the line segment from the pivot point to the center of mass. The force due to gravity perpendicular to this lever arm is mg*sin(θ), as stated in my first post. So, the torque should be L(mg\sin{θ}) where L = \sqrt{h^2 + w^2} and...

Homework Statement I'm wondering this for any object with moment of inertia I, but I'll ask this question for a rectangle for simplicity and I'm sure I can extend it to general objects. Say we have a rectangular object (with mass m, height h, and width L) that is attached to a wall by some...