Recent content by Lma12684

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    Static Equilbrium

    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.
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    Static Equilbrium

    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...
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    Static Equilbrium

    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...
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    Static Equilbrium

    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?
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    Static Equilbrium

    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?
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    Static Equilbrium

    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????
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    Static Equilbrium

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

    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 lenght that the window cleaner...
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    Conservation Laws in Rotational Motion Physics

    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
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    Conservation Laws in Rotational Motion Physics

    Ok, I recalculated and found: 1) 2 J 2) 4.99 J 3) 3.99 J
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    Conservation of Mechanical Energy

    Because they cancel??
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    Conservation Laws in Rotational Motion Physics

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

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