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|dw:1349552917635:dw|

The normal force, N\[N=425\]
The force along the horizontal, Fx, is\[F_x=142\cos(\theta)\]

So yes, you are correct.

no but the answer is .235

|dw:1349553164246:dw|

i know i am wrong somewhere

answer should be .235

|dw:1349553244013:dw|

ok how do u solve it

\[\sum_{F}^{?}=142\]

i am confused

|dw:1349553460977:dw|

tan theta

what is force of friction

instead of that can ii say 142cos60?

fx=71

i got i got ohhhhhhhh it is simple guys

|dw:1349554073776:dw|

so friction is 71 by finding the Fn=302

|dw:1349554155311:dw|

thx u for helpinggggggggggggg me

i have only one question how did u get the equation Fn=mg-fy

http://gyazo.com/cef28f38aa4b15271c0397dce1f96da1

i got u thx