Recent content by lichenguy

I made a mistake, all good now. =]

Umm, guys, i just did this using: 2Ffront - 2Frear = a(M + 4m), τ - FfrontR = Iα, FrearR = Iα, but it gave me ##a = \frac {2τ} {R(M+2m)}## instead of ##a = \frac {2τ} {R(M+6m)}## Is something missing?

Alright, finished, finally. Here are the results: $$v(t) = v_T (1-e^{ - \frac t τ})$$ $$x(t) = v_T τ(\frac t τ + e^{ - \frac t τ} -1)$$ Where ##v_T = \frac F b## and ##τ = \frac m b## On d: I first found that velocity is ##v = \frac {v_T} 2## at time ##t=ln(2)τ## Using ##W=ΔK+ΔE_{internal}##...

In this problem, if i compute ##τ=m/b## i get that ##τ=0.25## seconds. What fraction of ##v_T## does the block have when ##τ=0.25## seconds? Is that a bad question? Because it depends what ##v## is at the start, right?

So, ##R## is not an independent variable and that is why we can't use it? Using ##b## instead yields: ##v_T=F/b## And solving for dimension ##T## gives: ##m/b## But, what does this ##T## mean here? The time constant means dropping the initial value to ##1/e## percent or increasing the value to...

##b## has dimensions ##MT^{-1}## and together with ##v## i got the ##R=-\frac {LM} {T^2}## above. Yeah, i already did that. I guess i don't understand "You can't have dimensions of ##v_T## on both sides of the equation.". I understand what the time constant does now. I found a cool video on the...

Homework Statement A block of mass ##m = 1.00 kg## is being dragged through some viscous fluid by an external force ##F = 10.0 N##. The resistive force can be written as ##R = -bv##, where ##v## is the speed and ##b = 4.00 kg/s## is a phenomenological constant. You may ignore gravity (we...

Alright :D i didn't think of it hitting the wall, hehe. Thanks lad!

There will be need for an increased frictional force to stay up. If the frictional force doesn't increase it will fall over, i guess. I tried to find out if the rate of change of μ that we found earlier was enough to maintain equilibrium while moving the base back. I guess that's not what was...

Does it depend on the length of the ladder? Because i found something, but I'm not sure if it's right.

Ok, ty, i get the maximum frictional force thing now. On the e question, pulling it back will decrease the angle, but am guessing it won't change the angle in the same way as before? Before we moved the top, not the bottom.

A uniform beam of length L and mass m is inclined at an angle θ to the horizontal. Its upper end is connected to a wall by a rope, and its lower end rests on a rough, horizontal surface. The coefficient of static friction between the beam and surface is μs. Assume that the angle θ is such that...

Cool! :partytime: Thank you for all the help. :thumbup:

It's a contact force. Frictional force, i guess. Is it: τ - FfrontR = Iα?

Could it be FrearR = Iα? Is it the same at the front?