Recent content by clintau

Fcx is the T in the upper string. How do you get the T in the lower? Tupper-mg? gives app 69N. Answer should be 56.2 what is wrong?

how would you find the tensions in the strings?

so Fc=4*6^2/r=109.09 That gives the force in the x direction...What about the Y?

Y direction Fupper= T*cos(theta)+mg=0 Flower= T*cos(theta)-mg=0 yes no?

i don't know.

1. A 4.0kg object is attached to a vertical rod by 2 strings. Object roatates in a horizontal circle at a constand speed 6.00 m/s. Find the tention in the upper and lower string. The strings are 2 meters long. the distance between the attachment points for each string is 3 meters. 2...

thank your help

he asks for the equation of vertical motion y(t) and horizontal motion x(t) I don't know how to express that

Ymax=Yi+ViyT-4.9t^2 Ymax=19.6t-4.9t^2 Vf=Vi+at @ymax Vf=0 a=9.8(down) t=2.0=Tymax Ymax=0+19.6(2)-4.9(4) Ymax=19.6 Correct

if you have to use those equations in part a of the question why would my teacher ask for the equation of the motion in the x and y directions in part b?

In review of the post it appears that I didn't know the RELEVANT EQUATIONS. I did. I have found ymax to be 19.6 and tymax to be 2.0

xf=xi+vit+at2 xi=0 vi=5 a=? t=4sec (or 2*tymax) Xf=0+(5*4)+8a yes no...I still don't know where this gets me toward writing an equation for the motion.

do you mean: xf=xi+vit+1/2at^2 and vf=vi+at

no what are they?

No air resistance. The x component is 5 and y is 19.6. I don't know how to write the equation of the line. would it just be x=5.0? It doesn't seem that easy considering entire motion is parbolic? Would y=19.6 be correct for the motion in the y direction?