14 Nov '07 00:38

A 0.500 kg mass on a 3.20 meter string moves 2.5 radians/sec (1 revolution = 2 x pie x radians) in a horizontal circle. What angle does the string make with the vertical?

- Joined
- 15 Jun '06
- Moves
- 16334

- Joined
- 15 Jun '06
- Moves
- 16334

14 Nov '07 00:451 edit

heres another one*Originally posted by tomtom232***A 0.500 kg mass on a 3.20 meter string moves 2.5 radians/sec (1 revolution = 2 x pie x radians) in a horizontal circle. What angle does the string make with the vertical?**

A sled weighing 200N rests on a 15 degree incline, held in place by static friction. The coefficient of static friction is 0.5.*(a)*What is the magnitude of the normal force on the sled?*(b)*Whatis the magnitude of the static friction on the sled?*(c)*The sled is now pulled up the incline at a constant speed by a child. The child weighs 500N and pulls on the rope with a constant force of 100N. The rope makes an angle of 30 degrees with the incline and has negligible weight. What is the magnitude of the kinetic fricition force on the sled?*(d)*What is the coefficient of kinetic friction between the sled and the incline?- Joined
- 09 Nov '07
- Moves
- 140

14 Nov '07 07:40first question is simple conical pendulum stuff:

let tension=T, weight = mg, angular velocity = w, length of string l, angle x.

resolve forces vertically:

Tcos[x] = mg

horizontally:

Tsin[x] = m(w^2) x lsin[x]

so T = m(w^2)l

subsituting into the first equation:

cos[x] = g/(w^2)l

and so x = 1.1, about 61 degrees.- Joined
- 14 Feb '04
- Moves
- 28719

False berry14 Nov '07 14:43

I agree, but x here is the angle the rope makes with the vertical. The "dip" is the additive inverse, 90 - 61 = 29 degrees.*Originally posted by etotheipi***first question is simple conical pendulum stuff:**

let tension=T, weight = mg, angular velocity = w, length of string l, angle x.

resolve forces vertically:

Tcos[x] = mg

horizontally:

Tsin[x] = m(w^2) x lsin[x]

so T = m(w^2)l

subsituting into the first equation:

cos[x] = g/(w^2)l

and so x = 1.1, about 61 degrees.- Joined
- 09 Nov '07
- Moves
- 140

- Joined
- 09 Nov '07
- Moves
- 140

14 Nov '07 19:01for the second one:

let normal reaction force = R, frictional force up plane = F, coeff. of friction = m.

a)

resolving parallel to the plane:

R = 200cos[15] = 193N

b)

limiting equilibrium implies F = mR

so F = 97N

c)

the frictional force now acts down the plane, so

resolving along the plane:

F = 100cos[30] = 87N

d)

resolving parallel to the plane:

R + 100sin[30] = 200

so R = 150N

limiting equilibrium implies F = mR

so m = 0.58

Last part im not so sure about -- can you use limiting equilibrium in this case??- Joined
- 14 Feb '04
- Moves
- 28719

False berry- Joined
- 09 Nov '07
- Moves
- 140

14 Nov '07 22:23Wow, ignore my parts c and d from above as I have ignored a force ðŸ˜²

c)

frictional force now acts down the plane

resolving along plane:

F + 200sin[15] = 100cos[30]

so F = 35N

d)

resolving parallel to the plane:

R + 100sin[30] = 200cos[15]

so R = 143N

limiting equilibrium implies F = mR

so m = 0.24

I hope ðŸ™‚