Recent content by hanlon

Homework Statement A curve of radius 68m is banked for a design speed of 85km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve? Homework Equations 1) FNsin(theta) = m*v2/r 2) FNcos(theta) - mg = 0 3) tan(theta) =...

awesome got the answer, thanks for your help

ok I get it now, F = 2Tsin(theta) -mg

2ft = 735n?

equal to 0?

vertical force would be (75kg)(9.8m/s^2) = 735N right?

thats one of the things I am having trouble with, I don't know how to calculate FT. Is the vertical component the force of the dude hanging there, and if so what's the horizontal? what is the 27kN for? what's the factor of 10 for? srry for asking so many questions but I can't seem to wrap my...

[PLAIN]http://img175.imageshack.us/img175/7074/9999k.png like that right ehh Fg should be Force of the dude but w/e

copy and paste error srry 75kg

srry forgot about those just editted

Homework Statement Christian is making a Tyrolean traverse as shown in the figure. That is, he traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 22 away. The rope must sag sufficiently so it won't break. Assume the rope can provide...

Is my whole solution wrong, or just the forces I used. I understand now that I didn't use tension force, but is the way I derived the equation wrong with Fby = Fax

Homework Statement Determine a formula for acceleration of the system shown in Fig. 4-45 if the cord has a non-negligible mass mc. Specify in terms of lA and lB, the length of cord from the respective masses to the pulley ( The total cord length is l= lA + lB)...

ah, I got a email back from my teacher since its going downwards the velocity component is negative. woops

115 = 0 + (3.73m/s)*t + 4.9t2 since velocity and acceleration are moving in the same direction at this point then using the quadratic equation -37.3 (+/-)sqr(37.32 - 4(4.9)(-115))/ 2*4.9 which ended up 2.355102041 and some other number I didn't record.