I love manipulations like this one. Reminds me of the first time I saw Lagrange multipliers - seemingly making stuff more complicated until you realize there is no need to actually calculate the λ, or in this case D2. Going to try to use this one from now on :smile:
From time to time I hear about people coming up with creative/"non-mainstream" solutions to classical physics problems, whether by looking from a very different angle or using some unusual math that's unknown to anyone but that one slightly quirky professor from faculty of mathematics. However I...
Thats what happens when your first exposition to calculus doesn't come from a mathematician :biggrin: The consistency is lax at best, although I still enjoyed that teaching style.
Anyways, in that case, I'll jump into Rudin after I'm done and use Abbott as a supplementary if I get completely...
I'm using gap year to prepare for B.S. in electrical engineering. Currently I'm solving through Spivak's "Calculus", Lang's "Introduction to Linear Algebra" and Velleman's "How To Prove It." I have three books on analysis, Rudin's "Principles of Mathematical Analysis", Abbott's "Understanding...
Oh wow, I can just integrate F(t) as indefinite integral, can't I? I don't need the 0 and t limits because the integration constant I'll be getting will mean initial velocity, which is zero in this case.
Also, why is it wrong mathematically to sub θ for ωt? Is it because I'm supposed to be...
Homework Statement
Force represented by F(t)=20+t+5t^2 [N] acts upon the rim of a disk. How much time has to pass before the disk has angular velocity of 200 revs\sec? R and I are known.
Homework Equations
\begin{equation*}
\tau=I\alpha
\end{equation*}
\begin{equation*}
\tau=R \times F...
If there is no angular acceleration, then the net torque is zero; its used to overcome friction and mg, right?
That is embarrassing, but seems I fixed it
\begin{equation*}
r=b\sin\alpha-0.5a\cos\alpha
\end{equation*}
x and y components of T are Tcosα and Tsinα. As haru said, no acceleration, which yealds
\begin{equation*}
F+Tsin\alpha-mg=0
\end{equation*}
Which gives maximum static friction as
\begin{equation}
\mu(mg-Tsin\alpha)
\end{equation}
Consistent with Suraj, yay.
Then in x
\begin{equation*}...
I've got my r expression, seems to check out with random values if I check by graphing
\begin{equation*}
r=b\tan\alpha-\frac{a}{2}
\end{equation*}
If I'm doing it this way, where does the friction play into this? Is there "frictional torque" around the corner?
I don't see how can I make a right angle here between the force along the rope and the point, unless I'm imagining extension of a rope, in which case I'm lost at how I would calculate it only knowing the hypotenuse. What am I missing here?
:sorry:If I understand Nascent correctly, then I wasn't supposed to add sin component to the torque. I was under the assumption that displacement vector has to be perpendicular to the force, but now that I think about it, it simply makes the torque maximum. If I don't have to worry about 90°...
Ah, he meant torque, as moment of force :sorry:
So I have my equation to keep things stationary in x direction
\begin{equation*}
\mu mg=Tcos\alpha
\end{equation*}
And the torque
\begin{equation*}
\tau=bTsin\alpha
\end{equation*}
(Am I correct in saying r=b?)
Is torque in this case acting...
Okay, so if the body is not moving horizontally, the cosine component of the force T has to be equal to the force of friction, meaning
\begin{equation*}
\mu mg=Tcos\alpha
\end{equation*}
The moments of inertia of the rectangle at the corner has to be
\begin{equation*}
I=\frac{m(a^2+b^2)}{3}...
Homework Statement
A rectangular box is being pulled by a rope with a force T on a horizontal surface, with a friction coefficient μ. What is the minimum angle α at which it will lift of the surface? See attached image.
Homework Equations
\begin{equation*}
\tau = rF\sin\alpha
\end{equation*}...