Answer:
Explanation:
m1 = 3.77 kg (0, 0 )
m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)
m3 = 2.46181 kg (16.7024 cm, 0 cm )
Let x and y be the coordinates of centre of mass.
![x = \frac{m_{1}x_{1}+ m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bm_%7B1%7Dx_%7B1%7D%2B%20m_%7B2%7Dx_%7B2%7D%2Bm_%7B3%7Dx_%7B3%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%2Bm_%7B3%7D%7D)
![x = \frac{3.77\times 0+ 6.7106\times 5.72 + 2.46181\times 16.7024}{3.77+6.7106+2.46181}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B3.77%5Ctimes%200%2B%206.7106%5Ctimes%205.72%20%2B%202.46181%5Ctimes%2016.7024%7D%7B3.77%2B6.7106%2B2.46181%7D)
x = 6.1428 cm
![y = \frac{m_{1}y_{1}+ m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7Bm_%7B1%7Dy_%7B1%7D%2B%20m_%7B2%7Dy_%7B2%7D%2Bm_%7B3%7Dy_%7B3%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%2Bm_%7B3%7D%7D)
![y= \frac{3.77\times 0+ 6.7106\times 11.44 + 2.46181\times 0}{3.77+6.7106+2.46181}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B3.77%5Ctimes%200%2B%206.7106%5Ctimes%2011.44%20%2B%202.46181%5Ctimes%200%7D%7B3.77%2B6.7106%2B2.46181%7D)
y = 5.9316 cm
Yes because the tree would be laying on the ground still
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
![\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft](https://tex.z-dn.net/?f=%5CRightarrow%20y%3D%5Csqrt%7B20%5E2-5%5E2%7D%3D19.36%5C%20ft)
Also, at any instant t
![\Rightarrow l^2=x^2+y^2](https://tex.z-dn.net/?f=%5CRightarrow%20l%5E2%3Dx%5E2%2By%5E2)
differentiate w.r.t.
![\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s](https://tex.z-dn.net/?f=%5CRightarrow%200%3D2xv%2B2yv_y%5C%5C%5C%5C%5CRightarrow%20v_y%3D-%5Cdfrac%7Bx%7D%7By%7D%5Ctimes%20v%5C%5C%5C%5C%5CRightarrow%20v_y%3D-%5Cdfrac%7B5%7D%7B19.36%7D%5Ctimes%202%3D0.516%5C%20ft%2Fs)