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1000 sats \ 4 replies \ @SHA256man 2 Dec \ on: Brain Not Braining : Part 6 science
F net M = M * a = Tension - M * g
F net m = m * a = m * g * sin(ɵ) - Tension
adding the two equations: a(M+m) = g [m * sin(ɵ)] - g * M
final answer: a = g {[m * sin(ɵ)] - M}/(M+m)
good review in 11 min here;
write
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69 sats \ 3 replies \ @noknees OP 23h
excellent! Correct steps too!
Congo to @SHA256man for this weeks BNB 🥳!
Here is the detailed solution in case some don't understand:

Given:

  • Mass of box on incline: m
  • Incline angle: \theta
  • Hanging mass: M
  • Pulley: frictionless bearing but with moment of inertia I and radius R (to be included for rotational effects)
  • Rope: light and inextensible
  • The system is released from rest at t=0
  • Incline is frictionless
Okay so,
  • Let acceleration of box along the incline be $a$ downwards (positive direction along slope down).
  • The hanging mass moves vertically downward with acceleration a.
  • Let tensions on the box side of the rope and hanging mass side be T_1 and T_2 respectively.
  • Because the pulley has rotational inertia, T_1 \neq T_2.
m a = m g \sin \theta - T_1
(Box accelerates down the incline, gravity component downward minus tension force opposing motion.)
M a = M g - T_2
(Mass accelerates downwards, gravity downward minus tension upward.)
Since the rope unwinds without slipping, the angular acceleration \alpha of the pulley relates to linear acceleration a by:
\alpha = \frac{a}{R}
Torque \tau on pulley is caused by difference in tensions:
\tau = (T_2 - T_1) R = I \alpha = I \frac{a}{R}
Rearranged:
T_2 - T_1 = \frac{I}{R^2} a
We have 3 equations:
  1. m a = m g \sin \theta - T_1
  2. M a = M g - T_2
  3. T_2 - T_1 = \frac{I}{R^2} a
Rewrite tensions from (1) and (2):
T_1 = m g \sin \theta - m a
T_2 = M g - M a
Substitute into (3):
(M g - M a) - (m g \sin \theta - m a) = \frac{I}{R^2} a
Simplify:
M g - M a - m g \sin \theta + m a = \frac{I}{R^2} a
Group terms with a on right:
M g - m g \sin \theta = M a - m a + \frac{I}{R^2} a
M g - m g \sin \theta = a \left( M - m + \frac{I}{R^2} \right)
a = \frac{M g - m g \sin \theta}{M - m + \frac{I}{R^2}}
  • If pulley moment of inertia I = 0, this reduces to a simpler formula:
a = \frac{M g - m g \sin \theta}{M - m}
(This case may be problematic if M = m, implying infinite acceleration—showing the importance of pulley inertia.)
  • If the pulley has significant I, the denominator increases, reducing acceleration. :)
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0 sats \ 2 replies \ @SHA256man 23h
ooph, i have to review the torque mechanics a bit, the pulleys i solved intuitively and then wrote out the solution in a structured manner; thank u for the challenge and an opportunity to review some basics!
  • have u looked at SCIENCIA from wooden books?
  • how do u format the equations into standard scientific format? what quick online or github tool do u use for that?
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0 sats \ 0 replies \ @SimpleStacker 23h
how do u format the equations into standard scientific format? what quick online or github tool do u use for that?
Look up MathJax and LaTeX
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69 sats \ 0 replies \ @noknees OP 23h
looks worthy of reading, i must try it
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