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Thanks for all your help, but I couldn't figure the last part out, and the assignment was due today. I REALLY APPRECIATE ALL YOU HAVE DONE. Thank you.

Homework Statement The system at the right is in static equilbrium, and the string in the middle is exactly horizontal. Find a) tension T(1) b) tension T(2) c) tension T(3) d) angle thetas **I don't know if anyone can help me with this because I cannot upload the picture. Any help...

Ok here we go again: I really appreciate your help. F(fr)=F(w) F(fr)=mu F(n) =.4(117)(9.8) =458.64 N Thus, T(wall)=F(w)(L)(sin 65) =(458.64)(10)(sin65) =4156.69 Now, I am still lost on T(painter). 1. t=rFsintheta: here is the angle between the...

Here we go: T(wall)=Fw* L *sin theta =267.89*10*sin 65 =2427.90 T(ladder)=-22 (9.8) (10/2) (cos 65) =-455.58 T (painter)= mg (h) =(95)(-9.8)(h) =-931h 0=2427.90-455.58-931h x=2.104 meters Where did I go wrong?

Ok, here is what I found: T(wall)=2427.90 T(ladder)=-455.58 T(painter)=-931x 0=2427.90-455.58-931x x=2.104 meters up ladder How does it look?

I used the following formula: sum torque=F(w) * l (sin theta) - mg * L/2 cos theta=0 m=22kg THis is what I got: torque=267.89 * 10 * sin 65 - (22)(9.8) * (10/2) (cos 65) =4674.19 N What am I missing?

Was my attempt at solving anywhere close to what I am looking for? Thanks.

Thank You Again!

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 cloor with a coefficient of static friction equal to .40. What is the max length that the window cleaner...

Here is what I did: 2) KE=1/2MV(cm) + 1/2 I(cm)w^2 =1/2(2.5)(2.0) + 1/2(.0196)(204.08) =4.99 J 3) L=Iw =(.0196)(204.08) =3.99 J

Thank You~!

Ok, I recalculated and found: 1) 2 J 2) 4.99 J 3) 3.99 J

Because they cancel??

Did I forget the square on #2? Is that what you are saying? Thanks.

My apologies, they represent the radius. I didn't mean to use different symbols.

Homework Statement A uniform solid sphere, with diameter 28 cm and mass 2.5 kg, rolls without slipping on a horizontal surface, at constant speed of 2.0 m/s. 1) What is the rotational kinetic energy? 2) What is its total kinetic energy? 3) What is its angular momentum? Homework...

Homework Statement A massless string is wrapped around a solid cylinder as shown in the diagram. A block of mass 2kg hangs from a string. When released, the block falls a distance of 82 cm in 2.0 seconds. Calculate the mass using the conservation of mechanical energy. Homework...

Sorry, I just saw your post. Let me check it again. Thanks.

I think I figured it out. I had my I(cm) in the wrong place of the equation. I now found that I=.0762. THanks!

I found that I=.0366 kg/m. Now, I used I=I(cm) + m(d^2). .0366=I + (.44)(.3^2) I=-.003 I know I did something wrong. Did I use the wrong distance?

My mistake, I misunderstood you. The formula I found for a long uniform rod was I=1/12(M)(L^2). For L, so I consider it as 1 meter? Or do I use .5 m? I am not sure what you mean by parellel axis theorem. Thanks!

I am confused about the KE(cylinder) in the above equation. Can someone please explain this? Thanks.

Relavent formula: I=(sum of t)/angular acceleration

Should I assume that 30 cm is the middle of the rod? Thanks.

Homework Statement A meter stick of mass .44 kg rotates, in the horizontal plane, about a vertical axis passing thru the 30 cm mark. What is the moment of inertia of the stick? Treat it as a long uniform rod. Homework Equations I=1/2M(R^2) The Attempt at a Solution...