Реши пожалуйста

创建于：2025年12月24日

问题

IMG_20251224_160125.jpg

Реши пожалуйста

Python

python
from PIL import Image, ImageEnhance
img = Image.open('/mnt/data/IMG_20251224_160125.jpg')
img.size


(2797, 1771)

Python

python
import matplotlib.pyplot as plt, numpy as np, PIL
# rotate 90 deg clockwise? image seems sideways.
rot = img.rotate(90, expand=True)
rot.size


(1771, 2797)

Python

python
# crop central area with text for readability
import numpy as np
arr=np.array(rot)
arr.shape


(2797, 1771, 3)

Python

python
# Let's display quickly
plt.figure(figsize=(10,14))
plt.imshow(rot)
plt.axis('off')
plt.show()

Python

python
rot2 = img.rotate(-90, expand=True)  # rotate 90 counterclockwise
plt.figure(figsize=(10,14))
plt.imshow(rot2)
plt.axis('off')
plt.show()

Python

python
import numpy as np, matplotlib.pyplot as plt
# crop region with problem text (roughly middle top)
w,h=rot2.size
# crop around text lines
crop = rot2.crop((150, 300, w-150, 900))
# enhance contrast/sharpness
enh=ImageEnhance.Contrast(crop).enhance(2.5)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(12,6))
plt.imshow(enh)
plt.axis('off')
plt.show()

Python

python
crop = rot2.crop((80, 420, w-80, 1150))
enh=ImageEnhance.Contrast(crop).enhance(3.0)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(12,8))
plt.imshow(enh)
plt.axis('off')
plt.show()

思考

crop = rot2.crop((80, 420, w-80, 1150)) enh=ImageEnhance.Contrast(crop).enhance(3.0) enh=ImageEnhance.Sharpness(enh).enhance(2.0) plt.figure(figsize=(12,8)) plt.imshow(enh) plt.axis('off') plt.show()

text

![](https://files.chat01.ai/python-generations/files/file-D7HwVxtwngKEC9wFD7ETxU)



Thought for 1m 28s

回答

Чтобы стержень висел горизонтально, нить должна быть закреплена в центре масс системы (стержень + два шарика).

Пусть точка подвеса находится на расстоянии x от шарика m₁ (от его конца стержня). Тогда до шарика m₂ расстояние 30 − x, а центр стержня (его масса M) находится на отметке 15 см, значит до него расстояние x − 15 (точка подвеса будет правее середины, потому что справа тяжелее).

Условие равновесия моментов относительно точки подвеса:

m_1\cdot x + M\cdot (x-15)= m_2\cdot(30-x)

Подставим $m_1=10$ , $M=20$ , $m_2=20$ :

10x + 20(x-15)=20(30-x)

10x+20x-300 = 600-20x

50x = 900 \Rightarrow x = 18

Ответ: 18 см.

问题

IMG_20251224_160627.jpg

Реши пожалуйста

Python

python
from PIL import Image, ImageOps
img=Image.open('/mnt/data/IMG_20251224_160627.jpg')
img.size


(2523, 1788)

Python

python
import matplotlib.pyplot as plt, numpy as np
img_rot=img.rotate(90, expand=True)  # rotate to upright?
plt.figure(figsize=(10,7))
plt.imshow(img_rot)
plt.axis('off')
plt.show()

Python

python
img_rot=img.rotate(270, expand=True)
plt.figure(figsize=(10,7))
plt.imshow(img_rot)
plt.axis('off')
plt.show()

Python

python
import numpy as np, matplotlib.pyplot as plt
img_rot=img.rotate(270, expand=True)
# crop around diagram
crop=img_rot.crop((600,420,1900,950))
plt.figure(figsize=(12,4))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
crop=img_rot.crop((700,540,1700,820))
plt.figure(figsize=(12,4))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
import matplotlib.pyplot as plt
small=img_rot.resize((800,600))
plt.figure(figsize=(8,6))
plt.imshow(small)
plt.axis('off')
plt.show()

Python

python
w,h=img_rot.size
crop=img_rot.crop((500,900,1300,1450))
plt.figure(figsize=(12,6))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
import numpy as np
arr=np.array(crop)
# convert to grayscale
gray=arr.mean(axis=2)
# focus on region of ruler y between maybe 160-260 in crop
h,w=gray.shape
y1,y2=160,260
region=gray[y1:y2,:]
# compute column darkness: lower values means dark
col_mean=region.mean(axis=0)
# find local minima below threshold
threshold=np.percentile(col_mean,5)
mins=np.where(col_mean<threshold)[0]
# group into clusters
clusters=[]
current=[]
for x in mins:
    if not current or x==current[-1]+1:
        current.append(x)
    else:
        clusters.append(current)
        current=[x]
if current: clusters.append(current)
# get center of each cluster
centers=[int(np

(8, [334, 362, 374, 385, 388, 641, 678, 689])

Python

python
y1,y2=210,250
region=gray[y1:y2,:]
col_mean=region.mean(axis=0)
threshold=np.percentile(col_mean,3)  # darker
mins=np.where(col_mean<threshold)[0]
clusters=[]
current=[]
for x in mins:
    if not current or x==current[-1]+1:
        current.append(x)
    else:
        clusters.append(current); current=[x]
if current: clusters.append(current)
centers=[int(np.mean(c)) for c in clusters]
centers, len(centers)


([282, 334, 641, 679, 695], 5)

Python

python
import scipy.ndimage as ndi, numpy as np, math, statistics
# compute x gradient absolute
grad=np.abs(np.diff(region, axis=1))
# sum over y
score=grad.sum(axis=0)
# find peaks in score
# smooth
score_s=ndi.gaussian_filter1d(score, sigma=2)
# find peaks with prominence
from scipy.signal import find_peaks
peaks,_=find_peaks(score_s, prominence=np.percentile(score_s,95), distance=30)
peaks, len(peaks), score_s.max()


(array([282, 330, 643]), 3, 1270.8451128188112)

Python

python
peaks,_=find_peaks(score_s, prominence=np.percentile(score_s,85), distance=40)
peaks, len(peaks)


(array([282, 330, 413, 643]), 4)

Python

python
# enlarge crop for human inspection
big=crop.resize((crop.size[0]*2, crop.size[1]*2))
plt.figure(figsize=(14,6))
plt.imshow(big)
plt.axis('off')
plt.show()

Python

python
gray=np.array(crop.convert('L'))
# choose row y around 235
y=235
row=gray[y,:]
# threshold
dark=np.where(row<80)[0]
# cluster contiguous
clusters=[]
cur=[]
for x in dark:
    if not cur or x==cur[-1]+1:
        cur.append(x)
    else:
        clusters.append(cur); cur=[x]
if cur: clusters.append(cur)
# compute cluster centers and widths
centers=[int(np.mean(c)) for c in clusters]
widths=[len(c) for c in clusters]
list(zip(centers,widths))[:20], len(centers)


([(332, 1)], 1)

Python

python
# within ruler height approximate y from 180-270, compute dark count per x
ys=range(195,260)
mask=(gray[195:260,:]<120).astype(int)
col=mask.sum(axis=0)
# peaks where col> threshold
peaks=np.where(col>30)[0]
# cluster
clusters=[]; cur=[]
for x in peaks:
    if not cur or x==cur[-1]+1:
        cur.append(x)
    else:
        clusters.append(cur); cur=[x]
if cur: clusters.append(cur)
centers=[int(np.mean(c)) for c in clusters]
widths=[len(c) for c in clusters]
len(centers), list(zip(centers,widths))[:15]


(1, [(333, 6)])

Python

python
import matplotlib.pyplot as plt, numpy as np
# plot row average across ruler height for more robust
region=gray[185:270,:]
col_mean=region.mean(axis=0)
plt.figure(figsize=(12,3))
plt.plot(col_mean)
plt.ylim(0,255)
plt.show()

Python

python
# take narrow strip at top of ruler border near y=170-180
strip=gray[165:195,:]
col_min=strip.min(axis=0)  # minimum pixel value in strip per column
plt.figure(figsize=(12,3))
plt.plot(col_min)
plt.ylim(0,255)
plt.show()

Python

python
from scipy.signal import find_peaks
vals=255-col_min # invert so dips become peaks
vals_s=ndi.gaussian_filter1d(vals, sigma=1)
peaks,_=find_peaks(vals_s, height=40, distance=20)
peaks, len(peaks)


(array([335, 361, 388]), 3)

Python

python
strip=gray[190:270,:]
col_min=strip.min(axis=0)
vals=255-col_min
vals_s=ndi.gaussian_filter1d(vals, sigma=1)
peaks,_=find_peaks(vals_s, height=60, distance=40)
peaks, len(peaks)


(array([333, 384, 633, 678]), 4)

Python

python
peaks,_=find_peaks(vals_s, height=15, distance=40)
peaks, len(peaks), peaks[:20]


(array([251, 291, 333, 384, 426, 488, 542, 633, 678]),
 9,
 array([251, 291, 333, 384, 426, 488, 542, 633, 678]))

Python

python
big=crop.resize((crop.size[0]*4, crop.size[1]*4))
plt.figure(figsize=(16,8))
plt.imshow(big)
plt.axis('off')
plt.show()

Python

python
gray=np.array(crop.convert('L'))
# define region containing ruler (y bounds) by trial
y_top,y_bot=160,270
region=gray[y_top:y_bot,:]
col_min=region.min(axis=0)  # min in column
inv=255-col_min
inv_s=ndi.gaussian_filter1d(inv, sigma=1.5)
# find peaks
peaks,props=find_peaks(inv_s, height=30, distance=40)
peaks, props['peak_heights'][:10], len(peaks)


(array([290, 362, 405, 678]), array([ 49., 250.,  51., 253.]), 4)

Python

python
plt.figure(figsize=(12,3))
plt.plot(inv_s)
plt.scatter(peaks, inv_s[peaks], color='red')
plt.ylim(0,260)
plt.show()

Python

python
peaks,_=find_peaks(inv_s, height=35, distance=15)
len(peaks), peaks[:20]


(11, array([271, 290, 318, 334, 362, 388, 405, 427, 641, 678, 705]))

Python

python
import matplotlib.pyplot as plt
plt.figure(figsize=(12,4))
plt.imshow(crop)
for x in peaks:
    plt.axvline(x=x, color='red', linewidth=1)
plt.axis('off')
plt.show()

Python

python
gray=np.array(crop.convert('L'))
row_mean=gray.mean(axis=1)
# find y range where mean significantly less than background
import numpy as np
th=np.percentile(row_mean,10)
ys=np.where(row_mean<th)[0]
ys.min(), ys.max(), th


(294, 357, 230.035375)

Python

python
y_top,y_bot=300,360
region=gray[y_top:y_bot,:]
col_min=region.min(axis=0)
inv=255-col_min
inv_s=ndi.gaussian_filter1d(inv, sigma=1)
peaks,_=find_peaks(inv_s, height=40, distance=30)
peaks, len(peaks)


(array([262, 341, 376, 418, 497, 551, 645]), 7)

Python

python
plt.figure(figsize=(12,4))
plt.imshow(crop)
for x in peaks:
    plt.axvline(x=x, color='red', linewidth=1)
plt.axis('off')
plt.show()

Python

python
peaks,_=find_peaks(inv_s, height=20, distance=30)
peaks, len(peaks)


(array([192, 262, 341, 376, 418, 497, 551, 645, 704]), 9)

Python

python
ruler=crop.crop((240,260,760,410)) # focus
plt.figure(figsize=(12,3))
plt.imshow(ruler)
plt.axis('off')
plt.show()

Python

python
rgray=np.array(ruler.convert('L'))
# focus on interior excluding top/bottom border maybe y 50-110
region=rgray[30:120,:]
darkness=255-region
col_score=darkness.sum(axis=0)
col_s=ndi.gaussian_filter1d(col_score, sigma=1)
# find peaks with distance around segment width ~? use 50 maybe
peaks,_=find_peaks(col_s, prominence=np.percentile(col_s,85), distance=40)
peaks, len(peaks)


(array([ 30,  92, 153, 214, 275, 337, 398, 457]), 8)

Python

python
# get indices of highest scores
top=np.argsort(col_s)[-20:]
sorted(top)


[29,
 30,
 31,
 91,
 92,
 93,
 153,
 154,
 214,
 215,
 275,
 276,
 336,
 337,
 338,
 397,
 398,
 399,
 400,
 457]

Python

python
region=rgray[:, :]
darkness=255-region
col_score=darkness.sum(axis=0)
col_s=ndi.gaussian_filter1d(col_score, sigma=1)
peaks,_=find_peaks(col_s, prominence=np.percentile(col_s,85), distance=40)
peaks, len(peaks), ruler.size


(array([ 30,  92, 153, 214, 275, 337, 399, 457]), 8, (520, 150))

Python

python
col_s[-10:], col_s.max(), np.argmax(col_s)


(array([15, 26, 18,  6,  2,  5, 11, 16, 14, 14], dtype=uint64), 21377, 92)

Python

python
col_min=rgray.min(axis=0)
min_x=np.argmin(col_min)
col_min[min_x], min_x


(0, 0)

Python

python
xs=np.where(col_min<50)[0]
xs[:20], xs[-20:], len(xs)


(array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
        17, 18, 19]),
 array([448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460,
        461, 462, 463, 464, 465, 466, 467]),
 468)

Python

python
th=80
dark_frac=(rgray<th).mean(axis=0)
# find columns where dark_frac > 0.4
xs=np.where(dark_frac>0.3)[0]
# cluster
clusters=[];cur=[]
for x in xs:
    if not cur or x==cur[-1]+1:
        cur.append(x)
    else:
        clusters.append(cur);cur=[x]
if cur: clusters.append(cur)
centers=[int(np.mean(c)) for c in clusters]
widths=[len(c) for c in clusters]
len(centers), list(zip(centers,widths))[:15], centers[-5:]


(8,
 [(30, 4),
  (92, 5),
  (153, 4),
  (214, 4),
  (275, 4),
  (336, 4),
  (398, 4),
  (457, 3)],
 [214, 275, 336, 398, 457])

Python

python
plt.figure(figsize=(12,3))
plt.imshow(ruler)
for x in centers:
    plt.axvline(x=x, color='red', linewidth=1)
plt.axis('off')
plt.show()

Python

python
right=ruler.crop((480,0,520,150))
plt.figure(figsize=(3,3))
plt.imshow(right.resize((160,600)))
plt.axis('off')
plt.show()

Python

python
col_mean=rgray.mean(axis=0)
# find columns with mean < 245 (background white ~255)
cols=np.where(col_mean<245)[0]
cols.min(), cols.max(), ruler.size


(0, 468, (520, 150))

Python

python
gray_full=np.array(crop.convert('L'))
col_mean=gray_full.mean(axis=0)
cols=np.where(col_mean<250)[0]  # capture ruler and arrows perhaps
cols.min(), cols.max(), crop.size


(209, 707, (800, 550))

Python

python
row_mean=gray_full.mean(axis=1)
rows=np.where(row_mean<250)[0]
rows.min(), rows.max()


(147, 549)

Python

python
th=np.percentile(row_mean,5)
rows=np.where(row_mean<th)[0]
th, rows.min(), rows.max(), len(rows)


(228.6159375, 294, 357, 28)

Python

python
r_region=gray_full[294:357,209:707]
# compute dark fraction per column
dark_frac=(r_region<100).mean(axis=0)
# columns with dark_frac>0.2 indicate vertical lines
xs=np.where(dark_frac>0.2)[0]
clusters=[];cur=[]
for x in xs:
    if not cur or x==cur[-1]+1:
        cur.append(x)
    else:
        clusters.append(cur);cur=[x]
if cur: clusters.append(cur)
centers=[int(np.mean(c)) for c in clusters]
widths=[len(c) for c in clusters]
len(centers), list(zip(centers,widths))[:20]


(9,
 [(3, 5),
  (61, 6),
  (123, 5),
  (184, 5),
  (245, 5),
  (306, 6),
  (368, 7),
  (429, 6),
  (487, 6)])

思考

Identifying vertical line positions and boundary segments

We’ve detected 9 vertical line clusters, which likely mark both ends and 7 internal boundaries, giving us 8 segments. The left end is at x=3 and the right at x=487; segment width is roughly 60.5 px. The consecutive center distances are mostly uniform, confirming our count is accurate. Now, we’ll identify which boundary corresponds to F1 and F2 by finding the red arrow's position and determining their x coordinates relative to the ruler.

Thought for 9m 17s

回答

Чтобы линейка оставалась в покое, нужно выполнить два условия:

Сумма сил = 0
Сейчас обе силы сонаправлены: $F_1=5\text{ Н}$ и $F_2=1\text{ Н}$ .
Значит дополнительная сила должна быть противоположной и равной их сумме:

F_3 = F_1+F_2 = 6\text{ Н}

(направлена против $F_1$ и $F_2$ ).

Сумма моментов = 0
По рисунку силы приложены в делениях по 6 см:

$F_1$ стоит на границе после 2 делений $\Rightarrow x_1=2\cdot 6=12\text{ см}$
$F_2$ стоит на границе после 7 делений $\Rightarrow x_2=7\cdot 6=42\text{ см}$

Берём моменты относительно левого конца:

F_1 x_1 + F_2 x_2 = F_3 x_3

5\cdot 12 + 1\cdot 42 = 6\cdot x_3 \Rightarrow 60+42=6x_3 \Rightarrow x_3=17\text{ см}

Ответ:

дополнительная сила: 6 Н (в противоположную сторону)
приложить на расстоянии 17,0 см от левого конца.

问题

IMG_20251224_162226.jpg

Реши пожалуйста

Python

python
from PIL import Image, ImageEnhance
img=Image.open('/mnt/data/IMG_20251224_162226.jpg')
img.size


(2214, 1799)

Python

python
import matplotlib.pyplot as plt, numpy as np
img_rot=img.rotate(90, expand=True)
img_rot.size


(1799, 2214)

Python

python
plt.figure(figsize=(10,12))
plt.imshow(img_rot)
plt.axis('off')
plt.show()

Python

python
img_rot2=img_rot.rotate(180, expand=True)
plt.figure(figsize=(10,12))
plt.imshow(img_rot2)
plt.axis('off')
plt.show()

Python

python
import numpy as np, matplotlib.pyplot as plt
import PIL.ImageOps as ops
crop=img_rot2.crop((450,650,1350,1250))  # rough region
plt.figure(figsize=(10,6))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
# enlarge and show grid with markers
zoom=crop.resize((crop.size[0]*2,crop.size[1]*2))
plt.figure(figsize=(12,7))
plt.imshow(zoom)
plt.axis('off')
plt.show()

Python

python
import numpy as np, PIL.Image as Image
ruler = zoom.crop((0,250,1600,750))  # region around ruler
arr=np.array(ruler.convert('L'))
# find y range with ruler: pixels darker (border).
# Sum dark pixels per row.
dark = arr<80
row_sum = dark.sum(axis=1)
# rows with high dark count
ys = np.where(row_sum>100)[0]
ys[:10], ys[-10:], len(ys)


(array([], dtype=int64), array([], dtype=int64), 0)

Python

python
arr.min(), arr.max(), np.quantile(arr, [0.01,0.05,0.1,0.2])


(0, 255, array([166., 255., 255., 255.]))

Python

python
dark=arr<200
row_sum=dark.sum(axis=1)
ys=np.where(row_sum>200)[0]
ys[:20], ys[-20:], len(ys), row_sum.max()


(array([], dtype=int64), array([], dtype=int64), 0, 57)

Python

python
ys=np.where(row_sum>30)[0]
ys[:10], ys[-10:], len(ys), row_sum.max()


(array([24, 25, 26, 27, 28, 29, 30, 31, 32, 33]),
 array([315, 316, 317, 318, 319, 320, 334, 335, 336, 337]),
 59,
 57)

Python

python
plt.figure(figsize=(12,4))
plt.imshow(ruler)
plt.axis('on')
plt.show()

Python

python
ruler2 = zoom.crop((0,550,1600,1150))
plt.figure(figsize=(12,4))
plt.imshow(ruler2)
plt.axis('on')
plt.show()

Python

python
region = ruler2.crop((550,200,1550,370))
arr=np.array(region.convert('L'))
# detect dark pixels for borders, threshold 120 maybe
dark=arr<120
col_sum=dark.sum(axis=0)
# find peaks
import scipy.signal as signal, numpy as np
peaks,_=signal.find_peaks(col_sum, height=30, distance=20)
peaks[:20], len(peaks), col_sum.max()


(array([100, 202, 314, 419, 527, 637, 745, 853, 955]), 9, 141)

Python

python
# Use region2 bigger containing arrow and ruler
big = ruler2.crop((550,0,1550,500))  # includes arrow F1 and F2
arr=np.array(big)
# find red pixels: high R low G,B
R,G,B=arr[:,:,0],arr[:,:,1],arr[:,:,2]
red_mask=(R>150) & (G<120) & (B<120)
coords=np.column_stack(np.where(red_mask))
coords.shape


(2924, 2)

Python

python
# Separate arrow1 (up) and arrow2 (down) by x clustering
xs=coords[:,1]
# k-means 2 clusters
from sklearn.cluster import KMeans
km=KMeans(n_clusters=2, random_state=0).fit(xs.reshape(-1,1))
centers=sorted(km.cluster_centers_.flatten())
centers


[311.1820263308533, 852.3041631265935]

Python

python
centers_raw=km.cluster_centers_.flatten()
# map label->center
for lab in [0,1]:
    sel=coords[labels==lab]
    print(lab, centers_raw[lab], sel[:,0].min(), sel[:,0].max(), sel.shape[0])

思考

Determining F1 and F2 line indices

I’ve identified the y-ranges for F1 and F2: F1 at y=0-225 and F2 at y=339-429. This means F1 is around x=311 and F2 is around x=852. I need to match these x-values to vertical line peaks from the previously defined crop region. F1’s x-coordinate aligns with the third vertical line (index 2).

Balancing forces and calculating F3

F2 at x=852 aligns with the peak at 853, representing the 8th line. Since we have 8 segments, F2's boundary is 42 cm from the left. With F1 at 12 cm, F2 is at 42 cm. To achieve equilibrium, I need to calculate F3's force and location—F3 should be -4 N (downward) to cancel the net force. I’ll consider the moment conditions.

Calculating equilibrium and checking signs

I’m using a moment equation to solve for F3. First, I’m calculating moments around the left end: F1 provides a +512 moment, F2 gives a -142 moment, and F3 gives a -4*x3 moment. To balance, I set the sum of moments equal to 0 and solve for x3, which comes out to 4.5 cm. So, F3 is placed correctly. However, I’m double-checking the net force—right now we have 4 N upward, so F3 will need to be 4 N downward to balance things out.

Thought for 2m 47s

回答

Здесь силы противоположны: $F_1=5\text{ Н}$ вверх, $F_2=1\text{ Н}$ вниз.

1) Условие равновесия по силам

F_1 - F_2 - F_3 = 0 \Rightarrow F_3 = F_1 - F_2 = 5-1=4\text{ Н}

Дополнительная сила должна быть вниз.

2) Условие равновесия по моментам (относительно левого конца)

По рисунку деление = 6 см, всего 8 частей.

$F_1$ приложена на границе после 2 частей: $x_1=2\cdot6=12\text{ см}$
$F_2$ на границе после 7 частей: $x_2=7\cdot6=42\text{ см}$

Моменты (вверх — “плюс”, вниз — “минус”):

5\cdot 12 - 1\cdot 42 - 4\cdot x = 0

60-42 = 4x \Rightarrow 18=4x \Rightarrow x=4.5\text{ см}

✅ Ответ:

дополнительная сила: 4 Н
приложить на расстоянии 4,5 см от левого конца (вниз).

问题

IMG_20251224_163347.jpg

Реши пожалуйста

Python

python
from PIL import Image, ImageOps
img=Image.open('/mnt/data/IMG_20251224_163347.jpg')
img.size


(2446, 1805)

Python

python
import matplotlib.pyplot as plt, numpy as np, PIL
plt.figure(figsize=(10,7))
plt.imshow(img)
plt.axis('off')
plt.show()

Python

python
rot=img.rotate(90, expand=True)
plt.figure(figsize=(12,8))
plt.imshow(rot)
plt.axis('off')
plt.show()

Python

python
rot2=rot.rotate(180, expand=True)
plt.figure(figsize=(12,8))
plt.imshow(rot2)
plt.axis('off')
plt.show()

Python

python
import numpy as np
arr=np.array(rot2)
h,w=arr.shape[:2]
h,w


(2446, 1805)

Python

python
# crop central region
crop=rot2.crop((100, 200, 1700, 1800))
plt.figure(figsize=(12,8))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
# zoom text area around mass statement
text_crop=rot2.crop((120, 250, 1650, 550))
plt.figure(figsize=(12,4))
plt.imshow(text_crop)
plt.axis('off')
plt.show()

Python

python
text_crop2=rot2.crop((120, 350, 1700, 720))
plt.figure(figsize=(12,4))
plt.imshow(text_crop2)
plt.axis('off')
plt.show()

Python

python
L=100; l=25; M=1; r=2
x_s=L-l
# solve for a
import sympy as sp
a=sp.symbols('a', real=True)
sol=sp.solve(sp.Eq(r*(L-a/2-x_s), x_s-L/2), a)
sol


[25.0000000000000]

Python

python
a=25
x_cm=(M*L/2 + r*M*(a/2) + r*M*(L-a/2))/(M+2*r*M)
x_cm


50.0

思考

Verifying center of mass for symmetry

I find that the center of mass (CM) calculation matches with symmetry. After trying to compute the moments for the two cubes and the beam, I realized my earlier total was incorrect. By correctly calculating with the moments, I confirm that the CM is indeed at 50 cm, as expected, when the cubes are symmetrically placed.

Reconsidering support shift

After reevaluating, I realize the cubes are identical and symmetrically placed. The system is symmetric, so the center of mass (CM) is at 50 cm, as expected. Therefore, the support should be at the center of the beam, at 50 cm. The required shift is 25 cm, not 15 cm.

Verifying support movement logic

The question is asking about shifting the support to restore equilibrium after adding the second cube. Given the symmetry of the system with identical cubes at both ends, the center of mass should be at the center of the beam, requiring a 25 cm shift left. The symmetry supports this conclusion.

Reassessing cube positions and center of mass

If the cube is placed on the beam with one face flush to the end, the right cube would sit between L-a and L, and the left cube from 0 to a. This setup is symmetric, so CM ends up at the center, making the shift 25 cm. The initial equilibrium indicating a=25 validates this.

Finalizing shift determination and symmetry

With two cubes symmetrically placed, the system’s center of mass (CM) should be at 50 cm, indicating a 25 cm shift needed from the right to restore equilibrium. The question asks for this shift, not the new position. The answer is a 25 cm shift, confirming symmetry holds.

Clarifying support shift and torque equilibrium

They ask how much to move the support to restore balance, which implies it won't necessarily be at the center. Since the system is symmetric with identical cubes at the ends, the center of mass is at the midpoint, so a 25 cm shift from the initial 75 cm position to the new support at 50 cm is necessary. The shift can

Thought for 3m 25s

回答

Возьмём ось вдоль бруса от левого конца.

После того как на левый конец поставили такой же кубик, система стала симметричной: кубики одинаковые и стоят на концах, сам брус однородный ⇒ центр масс всей системы окажется ровно посередине бруса, то есть на

x=\frac{L}{2}=\frac{100}{2}=50\text{ см}

А подставка изначально была на расстоянии $l=25$ см от правого конца, то есть на координате

x_0 = 100-25=75\text{ см}

Значит, чтобы снова было равновесие, подставку нужно передвинуть на

\Delta x = 75-50 = 25\text{ см}

Ответ: 25 см.