Recent content by gianne671

so you agree with me? that my answer is correct?? 56.70N? how about the other equations? what did you get?

Homework Statement 1.)I can't get it why a friend of mine from this forum is getting 56.01N and me 56.70N 2.) I got this solution 50Ncos30-(0.02(20(9.8)-50sin30))/20, what did you get?? 3.) (0.4(20x9.8))/cos30+(0.4(sin30)) what did you get? Homework Equations 1)F= (0.1(50x9.8))/(cos25 -...

hello?

so the value which i get in the force needed to keep moving is 56.01 is the force app?? so.. 56.01cos 25-0.03*?/50 HELP!

here's the picture..

hey can i ask? what is the Fapp in the formula of acceleration? and why do I keep getting 56.70N instead of 56.01? i got this value of 49/0.8641=56.70N

Homework Statement What is the tension on the traction system?? i was given this picture.. the M= 10kg.. I got this picture,, the traction system is the figure that has a feet hanging. Homework Equations T=W=mg T=W=m2g The Attempt at a Solution T=W=10*9.8m/s =98N. T=W=10(2)...

what did you get? 56.70N? then for the acceleration the formula is = a=Fappcos/theta-ukFn/m am I right? what nos. did it give you(assuming that u already substituted for their value,, so that I can compare) and how can I solve for the time?

so the only problem is my first equation? then the rest 2nd,3rd,4th and Final equations are correct? :) can you please show me how to compute the time? needed to achieve a 5m/s?

I see.. so is this the right formula? Fcos/theta-us(fg-Fsin/theta=0 Fcos/theta-usFg+usFsin/theta=0 Fcos/theta+usFsin/theta=usFg F=us x Fg/cos/theta+ussin/theta so.. F= (0.1(50x9.8))/cos25+(0.03(sin25)) is that it? if not.. can you please illustrate the formula to be used for A) B) & C)...

tnx! so if I am not mistaken.. Fappcos/theta-f=maFn+Fappsin/theta=Fappcos/theta-u.. then.. Fappcos/theta-mukFn=maFn=Fg-Fappsin/theta that gives me.. a=Fappcos/theta-ukFn/m am i right?

Im trying to draw it.. but i can't still get it.. :( I am sorry for being a NOOB.. (teary eyes)

Fx= kinetic friction of 0.03 and static of 0.01 Fy= normal force and the pull of Earth on the block and 25 degree? that's all I can think of ..

ohh isee! I think there are two forces. it's the normal force and the g.. :) is that it?

the force acting on the block was named force "F" directed at 25 degree below the horizontal..