Recent content by ledphones

Thank you!

could B d/dt (x1 -x2) be equal to B (dot{x}1 - dot{x} 2)? Thank you! dot{x} is supposed to be a "dot" over a x. Just a formatting problem

Homework Statement A 3/4-in.-diameter rod made of the same material as rods AC and AD in the truss shown was tested to failure and an ultimate load of 29 kips was recorded. Using a factor of safety of 3.0, determine the required diameter (a) of rod AC, (b) of rod AD. Homework Equations...

Homework Statement a)Evaluate ∫∫∫E dV, where E is the solid enclosed by the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2 =1. Use the transformation x=au, y=bv, z=cw. b)If the solid in the above has density k find the moment of inertia about the z-axis. Homework Equations ∅=phi The Attempt...

Homework Statement A student is standing on a ladder as shown in the figure to the right. Each leg of the ladder is 2.6 m long and is hidged at point C. The tie-rod (BD) attached halfway up and is 0.79 m long. The student is standing at a spot 1.95 m along the leg and her weight is 510 N...

Homework Statement The above figure shows a uniform iron beam of mass 253 kg and length L = 3 m. The cable holding the beam in place can take a tension of 1300 N before it breaks. Use the Young's Modulus for steel to be 200e9 N/m2. (You may ignore the small mass of the cable in this...

.0044945452(from the following expressions:) 1/2Iw^2 I=(2*MR^2+mR^2)

Homework Statement A refrigerator is approximately a uniform parallelepiped h = 8 ft tall, w = 3 ft wide, and d = 2 ft deep. It sits upright on a truck with its 3 ft dimension in the direction of travel. Assume that the refrigerator cannot slide on the truck and that its mass is 130 kg. For...

Ok so now I tried 1/2Mv^2/r^2=Mgh h=.5(2.1)(3.9^2)/(.315^2*9.81*2.1) this was wrong :(

ok so i tried: 1/2(MR^2+MR^2+mr^2)w^2=mgh h=(.5(2*2.1*.351^2+.050*.351^2)*.146^2)/(.05*9.8)

i tried just M now. and that was wrong as well. as for my reason for adding those two i though you had to add the mass of the wad of gum plus one of the balls

so you'd know the potential energy once the wad hits is (m+M)gh. so i'd set that equal to 1/2 Iw^2 and solve for h. but this was a value of .0002133 which is much to small.

how would you use that to solve for the angle? would you have K+U=K+U 1/2 I w^2 + 0 = 0 + ? what would the potential energy be here? and how could you relate it to the angle (or distance traveled?) would it be through a kinematic equation?

Homework Statement Two 2.1 kg balls are attached to the ends of a thin rod of negligible mass, 63 cm in length. The rod is free to rotate in a vertical plane about a horizontal axis through its center. With the rod initially horizontal as shown, a 50 gm wad of wet putty drops onto one of the...

Thank you so much! That was exactly what I was doing wrong. Mass was the mass of the disk with the hole. Not the disk without the hole.