Recent content by mmravunac

So i think I would have to set the equation to Mb is Beam mass which is 0 (?)T*4d - Mbg*2.5d - 294N*4d*cos(θ) = 0 which is set to T*4d = 294N*4d*cos Dividing each side by 4d gives us T= 294Ncos(θ) is this right or can I do more? EDIT: on my graphing calculation when i type in 294cos(theta)...

http://s3.amazonaws.com/answer-board-image/ce8f6cfa-260b-4780-bb33-e34383a726e2.png This is the link to the diagram

1. The sign Ye Olde Tavern is supported by a horizontal beam (A-C) and a cable (B-C). The sign has a weight of W= 294 N. a) Take (Summate) the torques of the system about point A then find the tension in the cable. b) Find the vertical force on the wall attachment at point A by taking torque...

I think I have it now. Xcm= (1.3m1*Lo) / 2.3m1 Xcm=.57Lo (which makes this the distance from the smaller person) ? And for the second part, would this mean i set it up like Xcm=(1.3m1*.5L) / 2.3m1

So I cannot cancel out the m2. I am not positive where to go on from Xcm=(m2X2) / (m1+m2) Would this be my final equation or is there more I can do?

This is that part of the equation that has been tricking me up. I want to say that Xcm=(m2X2) / (m1+m2) So I believe I can take out m2 from each side of the equation giving me Xcm=(X2)/(m1) (?)

I believe X1 would be 0 and that would make X2 the whole length making it equal to Lo?

I am not sure of what exactly you are asking but center of mass is equal to Xcm= (m1X1 + m2X2) / (m1 + m2)

Suppose two friends, with masses m2 = 1.3 m1, are on a perfectly smooth, frictionless, frozen lake. They are both holding the end of a rope of length Lo . a. Find the position of the center of mass, in terms of Lo , from the smaller person. b. If the two pull half of the rope in such...