Recent content by gmichel395

Ah so six. And the answer is 92kg?

I count six sections plus the one the girl is holding so seven. 7*150 divided by g gives approximately 110kg?

Homework Statement "Using the pulley system shown to lift the heavy dresser, the person is applying a force of 150N to the end of the rope. Which of the following is closest to the maximum mass of the dresser that can be lifted with this force? [Ignore weight & friction from pulleys/rope.]" A)...

Ah, θ=90°=1.57rad. t=Δθ/ω. t=.48s. Then √(652+652) divided by .48s gives the 190cm/s2. Thank you!

So if I subtract vi from vf, I get <65,-65>. To find time I used the formula acentripetal=v2/r=ω2r, and solved for ω by plugging in 65 cm/s as v and 20 cm as r and found ω to be 3.25rad/s. I converted this to rev/s and multiplied by 1/.25rev (since 1/4 the way around track) and got time to be...

Would vi=<0,65> and vf=<65,0>?

Homework Statement "Assume that a train is traveling at 65 cm/s, and the track has radius 20cm. What is the magnitude of the train's average acceleration as the train goes from the tree marked A to the next tree B, located exactly 1/4 of the way around the track from tree A? [Hint: This...

Ah... The force that the hand exerts on the rope must be equal to the force the rope exerts on the hand. The force that the rope exerts on the mass must be equal to the force that the mass exerts on the rope. But the force that the hand exerts on the rope does not have to equal the force that...

Using the original problem with frictionless surface: Rope ∑Fx=FpullONrope-FmassONrope=ma Mass ∑Fx=FropeONmass=ma So the rope has a very small mass, and it is definitely accelerating. The only way for there to be acceleration is if FmassONrope is less than FpullONrope but how can that be since...

Trying to do this, but not sure what you mean by a "horizontal force balance equation."