Recent content by overdrive7

That part I don't understand. It's hard to believe that you can hold it at any angle and have it stay there. Are you saying if I held the beam almost vertical, maybe at an angle 10 degrees to the vertical, it would stay that way? There has to be some relation between the weights on the ends and...

It makes sense on paper that the angles will just cancel out, but then what changes need to be made to the system to obtain an angle of 45 degrees, from 30 degrees?

No problem at all, I understand it better now thanks to you. The only part I don't understand is how a change in the weight would affect the angle. Since you said the angle doesn't really matter, how would the change in angle matter? How does the 15 degree change to 45 degrees relate to the...

I see, makes sense. But how about the (w x 5)? I thought it would be w x 4, for the distance from w to the hanging point.

how did you get (140 x 1) + (w x 5)? Is there a reason you put all the weight of the beam to the left?

I apologize, the length of the beam is 6m. The 100-N weight is 2m to the right of the hanging point, and the other weight "w" is on the left, 4m from the hanging point.

Homework Statement a. A uniform beam is hanging from a poing 1.00 m to the right of its center. The beam weights 140 N and makes an angle of 30° with the vertical. At the right-hand end of the beam, a 100-N weight is hung; an unknown weight w hangs at the left end. If the system is in...

Thank you! I will use that and find the integral from infinity to the surface of the earth, since it says "at large distances, the force is 0", so the infinity would cancel out and leave the KE when it hits the surface.. I hope that's correct.

Thank you for the reply. By "big G formula", are you referring to the Gm1m2/r^2 formula? If so, is there a way to do it without the big G? I don't think I'm supposed to use the gravitational constant, because that is chapters away from where we are right now in our textbook (ch13 vs ch6).

Homework Statement The gravitational pull of the Earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the Earth's surface this force is equal to the object's normal weight mg, where g = 9.8m/s^2, and at large distances, the...