Qfwfq Posted April 11, 2006 Report Share Posted April 11, 2006 Alright, I'll change the questions:Let coefficent = a Net force on body = 3-3a=...What is the red 3, and what is the blue 3? Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 11, 2006 Report Share Posted April 11, 2006 The blue three is the original force. The 3a is the Friction the 3 there is due to Fr=uR and R (the normal reaction) in this case is 3 Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 11, 2006 Report Share Posted April 11, 2006 In this case I have taken a to be the coefficient of friction times g the acceleration due to gravity Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 12, 2006 Report Share Posted April 12, 2006 Now that sorts thing out! In this case I have taken a to be the coefficient of friction times g the acceleration due to gravityThat isn't the way it is defined, it wouldn't be the ratio of force to force, it would be the mass to force ratio... an acceleration... not a property of the two surfaces in contact. Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 12, 2006 Report Share Posted April 12, 2006 Sorry about that. I did the working out on a piece of paper that I loist and had to type up from memory and forgot I had lumped u and g together. Quote Link to comment Share on other sites More sharing options...
n1ckz0r Posted September 29, 2006 Report Share Posted September 29, 2006 That's all well and good for the two blocks. . . However everyone left the poor chicken out to dry (pun intended). Here's how you work out that one. Because the chicken isn't moving the net force=0Therefore, when broken into components, the X net force is 0, as is the Y net force. X:Net Force=Force 1 + Force 20=T1cos 30 - T2cos 45 (side note: the cosines were introduced due to the x component. T2 is negative because of the opposite direction.)T2cos 45=T1cos 30T2cos 45=T1 cos 30 The above is equation 1. Y:Net Force=Force 1 + Force 2 + weight0=T1sin 30 + T2sin 45 - mgFortunately, we have a value for T1, noted in equation 1mg=(T2cos 45/cos 30)sin 30 + T2sin 45mg=T2cos 45tan 30 + T2sin 45 (side note:sin30/cos 30=tan 30)mg=T2(cos 45tan 30 + sin 45)(2kg)(9.8m/s^2)/[(0.7071)(0.5774) + (0.7071)=T219.6N/1.115=T218N=T2 (Rounded for significant digits) As though by miraculous circumstances, we now have a value for T2, to substitute into our equation 1. 18Ncos 45/cos 30=T115N=T1 Therefore, the wire must be able to support at least 18 Newtons of tension for the chicken to remain there, rather than fall to its gruesome death. Quote Link to comment Share on other sites More sharing options...
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