Brain Not Braining : Part 6 \ stacker news ~science

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173 sats \ 9 comments \ @noknees 2 Dec science

Guess we're all fine so let's get down to business shall we?

Problem:

A box of mass m is placed on a frictionless incline at an angle $θ$ to the horizontal. The box is connected by a light, inextensible rope over a frictionless pulley at the top of the incline to a hanging mass M.

At time t=0, the system is released from rest.
Assume the rope does not slip on the pulley.
The incline and pulley apparatus are fixed.

Question: What is the acceleration of the box just after release?

(i know the image looks weird, but it's AI generated and I hope you get the idea)

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4 replies \ @SHA256man 2 Dec

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;

69 sats \ 3 replies \ @noknees OP 3 Dec

excellent! Correct steps too!

Congo to @SHA256man for this weeks BNB 🥳!

Here is the detailed solution in case some don't understand:

Given:Given:

Mass of box on incline: $m$
Incline angle: $θ$
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, $T_{1} \neq = T_{2}$ .

ma = m g sin θ - 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 $α$ of the pulley relates to linear acceleration $a$ by:

α = \frac{a}{R}

Torque $τ$ on pulley is caused by difference in tensions:

τ = (T_{2} - T_{1}) R = I α = I \frac{a}{R}

Rearranged:

T_{2} - T_{1} = \frac{I}{R ^{2}} a

We have 3 equations:

$ma = m g sin θ - T_{1}$
$M a = M g - T_{2}$
$T_{2} - T_{1} = \frac{I}{R ^{2}} a$

Rewrite tensions from (1) and (2):

T_{1} = m g sin θ - ma

T_{2} = M g - M a

Substitute into (3):

(M g - M a) - (m g sin θ - ma) = \frac{I}{R ^{2}} a

Simplify:

M g - M a - m g sin θ + ma = \frac{I}{R ^{2}} a

Group terms with $a$ on right:

M g - m g sin θ = M a - ma + \frac{I}{R ^{2}} a

M g - m g sin θ = a (M - m + \frac{I}{R ^{2}})

a = \frac{M g - m g sin θ}{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 θ}{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. :)

0 sats \ 2 replies \ @SHA256man 3 Dec

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?

69 sats \ 0 replies \ @noknees OP 3 Dec

looks worthy of reading, i must try it

0 sats \ 0 replies \ @SimpleStacker 3 Dec

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

10 sats \ 1 reply \ @Scroogey 2 Dec

The masses and the pulley are irrelevant when you ask about the acceleration?

$s in (θ) \times 9.81 \frac{m}{s ^{2}}$

0 sats \ 0 replies \ @noknees OP 2 Dec

no but the pulley is a frictionless bearing but with moment of inertia I and radius R (to be included for rotational effects)

10 sats \ 1 reply \ @SimpleStacker 2 Dec

Haha that image itself is like a MC Escher painting

0 sats \ 0 replies \ @noknees OP 2 Dec

lol true