Hey, I just have a quick question. I understand that the work done by a spring is considered as the following; Ws=-.5kx^2. What if the spring has weight to it? Would the work done by the spring be Ws=-(.5kx^2)-m(spring)g? Does the weight get added or is it not even considered in that equation...
I understand the idea of using torques. We're trying to separate the lifts into two different problems. Main problem being that the distance we'd use in any formula is relative to what? the center of the beam?
Having a problem figuring this out; it appears extremely easy but I would understand if rotational motion comes into play. I'm just not sure where.
A cylinder of mass 2kg is rolled up an incline by means of a string arranged as shown in the figure. (incline with angle theta and a cylinder...
Not sure about this one, any help?
Two men want to carry a wooden beam weighing 200 lbs which is 20 feet long. Both men are equal height however one is stronger than the other and wishes to bear 50% more the weight. How far from the end of the beam should both of the men be to make this possible?
Not sure about this one, any help?
Two men want to carry a wooden beam weighing 200 lbs which is 20 feet long. Both men are equal height however one is stronger than the other and wishes to bear 50% more the weight. How far from the end of the beam should both of the men be to make this possible?