Recent content by Lma12684

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.

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...