Recent content by rcwha

yeah it makes sense now...thank you guys

...that the normal force is zero?

so i don't have to wrry about the horizontal forces? i thought the compressive forces meant the horizontal forces...anyways this is what i did...i found friction to = 60.45 by multiplying 95.5,which I am asumming is the normal force time .633 (coef. friction). So the board is slipping down with...

the sandwhich of boards lies in the vertical plain...i added an http://

The board sandwiched between two other boards weighs 95.5 N. If the coefficient of friction between the boards is .663, what must be the magnitude of the comrssion forces (assume horizontal acting on both sides of the center board to keep it from slipping? I have no idea how to do this...

like i said earlier, i got (3.6,13.6) as my final answer from using tekken's method...did i get it right?

k i followed ur steps and got (3.6,13.6)...can somebody please confirm my answer...also, when finding the (1/5th) distance between two points could you use (x1+x2)/5, y1+y2/5...i tried this but i got conflicting numbers compared to doing it the other way...

im trying to figure out the angle of two these vector problems...ive found the magnitude but I am unsure which angle i need to solve for question 1) A man lost in a maze makes 3 consecutive displacements so that at the end of his travel heis right back where he started. The first...

k...so i have Ax = 30 Ay= -20 Bx = 60 By = 80 half the distance from A to B would be adding (60/2) to Ax and (80/2) to Ay...so now I am at (60,20) Cx = -10 Cy = -10 half the distance from (60,20) to C would be adding (-10/2) to 60 and (-10/2) to 20...so now I am at...

Long John Silver, a pirate, has buried his treasure on an island with five trees located at the following points: A (30.0 m, -20.0 m), B (60.0 m, 80.0 m), C (-10.0 m, -10.0 m), D (40.0 m, -30.0 m), and E (-70.0 m, 60.0 m). All of the points are measured relative to some origin. Long John's map...

is it V^2=Vyo^2 + 2ay a= -9.8 im assuming final velocity= 0 b/c that's what v is going to = at its max height thank you...i ended up with the right answer

another 2 dimensional problem :( The best leaper in the animal kingdom is the puma, which can jump to a height of 11.5ft (3.51m) when leaving the ground at an angle of 43 degress. With what speed, must the animal leave the ground to reach that height? at first i thought i could find the...

thanks a lot everybody...i really appreciate the help

i just need help simplifying the problem and solving for t...can somebody please just show me how one would get 42 as the initial velocity... thank you

yeah, so do i make B= .57vt? if so, then ill get...-.57+or- Square root of(.3249v^2)-411.6) / (-9.8) ..i wouldn't know what to do from here...