minoupak Posted September 25, 2006 Report Share Posted September 25, 2006 Aghhh....i think I've finally crashed...i can't seem to figure a problem out...actually two please anyone help? 1. A 5.0 kg block is sent sliding up a plane inclined at 37 degrees while horizontal force F of magnitude 50N acts on it. The coefficient of kinetic friction is .30. a) magnitude of the block's acceleration Ok i know the direction of the acceleration is down the incline...but i'm really stuck on the equation to find the acceleration.... 2. A 1.34kg ball is connected by means of two massless strings, each of length L=1.7m to a vertical, rotating rod. The strings are tied to the rod with separation of d=1.70 m and are taut. The tension in the upper string is 35N. a) the tension in the lower string, :) the magnitude of the net string force on the ball c) speed of the ball I can get c, if i can get a)... Quote Link to comment Share on other sites More sharing options...
minoupak Posted September 25, 2006 Author Report Share Posted September 25, 2006 just to let everyone know that i'm not shamming....lol this is what i have for the first problem:@=theta..U=coefficient of kinetic friction in the x-dirFCos@-WSin@+UN=ma in the y-dirN-FSin@+WCos@=maN=FSin@+WCos@ So the equation by substituting N would be: FCos@-WSin@+U(FSin@+WCos@)=ma and acceleration according to this would just be all that divided by m....but something must be off bcuz that's not the correct answer...:) btw, the acceleration should be 2.1 m/s*s please someone tell me what is wrong with my equation? lol i'm desperate to know what i don't understand Quote Link to comment Share on other sites More sharing options...
max4236 Posted September 25, 2006 Report Share Posted September 25, 2006 The force of friction (UN) is in the negative x direction so you need to change the x-dir equation to FCos@-WSin@-UN=ma There's no acceleration in the y direction so the equation isN-FSin@-Wcos@=0N=FSin@+WCos@ So you getFCos@-WSin@-U(FSin@+WCos@)=maa=1/5.0*(50Cos37-49Sin37-.3(50Sin37+49cos37))a=-2.12.1 m/s^2 going down the incline I haven't looked at the 2nd question yet, just a sec. Quote Link to comment Share on other sites More sharing options...
max4236 Posted September 25, 2006 Report Share Posted September 25, 2006 Ok, the spinning string and ball system forms an equilateral triangle with the vertical rotating rod. If you set up the force equations like before, let T1=tension on the top string=35N and T2=tension on the bottom string, then x-dir: F=T1cos30+T2cos30=ma=m(v^2/r) (r=LCos30=1.7cos30=1.5m)y-dir: T1sin30-T2sin30-W=0 (W=mg=1.43kg*9.8m/s^2)=14N) T2=(T1sin30-W)/sin30=7NF=36N net string force on the ballv=sqr(F*r/m)=6.1m/s speed of the ball Have fun :) Quote Link to comment Share on other sites More sharing options...
minoupak Posted September 25, 2006 Author Report Share Posted September 25, 2006 haha yeah i eventually figured out that coefficient of friction was negative. Thanks for the second problem! I didn't know where to start the problem from. That helps alot Quote Link to comment Share on other sites More sharing options...
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