Recent content by wizzle

Luckily my boyfriend's sister, a civil engineering genius, helped me through this. Thanks everyone for your help along the way!

yay! That's good news. I'm submitting it for an assignment so I'm glad to hear it! I'll let you know whether it's right when I get it back :) Thanks for helping!

Hi rl.bhat, the answer I came up with x=9.41 m on the 10 m ladder. What do you think? Thanks! -Lauren

Hmm..ok so being one who doesn't often hang out on ladders, I'm trying to figure out whether 65 degrees is enough to prevent slipping. Is it possible that 9.41 m would be a reasonable answer? I came out with the same answer after trying it multiple times, so I can't see where I could be going...

Hi! Thanks a lot for attaching that link, it really helped. I think I've got it this time, since I'm using the right coordinates - (11x56)+(2x48) / (56+48) = 6.85 cm I appreciate your time srvs!

Ahh...thank you. Since I separated the two at y=4 cm, then the upper rectangle is 14 cm high. Should I have put (7x56)+(2x48) / (56+48) = 4.69 cm for the y coordinate? Thanks for your help!

Homework Statement A carpenter's square has the shape of an "L," as shown at the right. Find the coordinates of its center of mass, assuming it to be made of uniform, material. (Hint: divide the L-shape into two rectangles.) Homework Equations MaXa+MbXb/(Ma + Mb) The...

Homework Statement A window cleaner of mass 95 kg places a 22kg ladder against a frictionless wall at a angle 65 degrees with the horizontal. The ladder is 10 m long and rests on a wet floor with a coefficient of static friction equal to .40. What is the max length that the window cleaner can...

Great! I calculated I through the centre to be (1/12)ML^2 = (1/12)(.44 kg)(1 m)^2 = 3.67 x 10^-2 kg*m^2. I then used Iz = Icom + Md^2 = (3.67 x 10^-2 kg*m^2)+(.44 kg)(.2 m)^2 = 5.43 x 10^-2 kg*m^2. I don't know how to gauge whether that is a reasonable answer. Any chance you can spot a...

Hmm...the parallel axis theorem hey? Unfortunately I don't know what that is, but I'm looking it up. Are you referring to Iz=Icom+Md^2?

Moment of inertia of a rod: axis not through the centre!? Homework Statement A meter stick of mass 0.44 kg rotates, in the horizontal plane, about a vertical axis passing through the 30 cm mark. What is the moment of inertia of the stick? (Treat it as a long uniform rod) Homework Equations...

Thanks for not giving up Tim. I used m and M as you suggested: mv+Mv=mV+MV (mv-MV)/m=V Now I substituted V, the final velocity of the small ball, into the following equation: mgh=.5mV^2+.5MV^2 mgh=.5m(mv-MV/m)^2+.5MV^2 mgh=.5m((mv)^2-2mvMV+(MV)^2)/m^2+.5MV^2 I multiplied the .5MV^2 term by...

I just had a thought...I don't think my answer could be right because in order for the kinetic energy just after the collision to be equal to the kinetic energy pre-collision, the velocities I came up with couldn't be right. My answer would mean that EK final would be greater than EK initial...

Hi Tim, I think I may have got the answer to the first part. After countless re-calculations, I finally have V'b to equal 3.715 m/s, which was the velocity of the smaller ball just before the collision, and V'a is now equal to 0 m/s, so all the kinetic energy was transferred to the larger...

sorry Tim, I don't mean to be wasting your time. my calculations seem to be sloppy, I'm going over them again.