Frictional forces are exerting force against the basketball, causing it to stop.
P2o3 is <span>diphosphorus trioxide.
Add water and you get phosphoric acid</span>
I believe it is true . Somebody correct me if I’m wrong!
Answer:
Explanation:
Given:
length of ladder ![r_L = 14m](https://tex.z-dn.net/?f=r_L%20%3D%2014m)
weight of ladder ![F_L = 490N](https://tex.z-dn.net/?f=F_L%20%3D%20490N)
position of firefighter ![r_F = 3.8m](https://tex.z-dn.net/?f=r_F%20%3D%203.8m)
weight of firefighter ![F_F = 820N](https://tex.z-dn.net/?f=F_F%20%3D%20820N)
angle of ladder ![\alpha = 63](https://tex.z-dn.net/?f=%5Calpha%20%3D%2063)
Unknown:
force of the wall on the ladder ![F_W](https://tex.z-dn.net/?f=F_W)
force of friction on base of ladder ![F_R](https://tex.z-dn.net/?f=F_R)
normal force on base of ladder ![F_N](https://tex.z-dn.net/?f=F_N)
From the free body diagram of the sketch you get 3 equations:
![F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0](https://tex.z-dn.net/?f=F_x%20%3D%20ma_x%20%3D%20F_W%20-%20F_R%20%3D%200%5C%5C%20F_y%20%3D%20ma_y%20%3D%20F_N%20-%20F_F%20-%20F_L%20%3D%200%5C%5C%20%5Ctau%20_P%20%3D%20%5Coverrightarrow%7Br%7D%20%5Ctimes%20%5Coverrightarrow%7BF%7D%20%3D%20r_FF_Fcos%5Calpha%20%2B%20%5Cfrac%7B1%7D%7B2%7Dr_LF_Lcos%5Calpha%20-%20r_LF_Wsin%5Calpha%20%3D%200)
Solving the equations gives:
![F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}](https://tex.z-dn.net/?f=F_W%20%3D%20F_R%5C%5C%20F_N%20%3D%20F_F%20%2B%20F_L%5C%5C%20F_W%20%3D%20%5Cfrac%7Br_FF_F%20%2B%200.5r_LF_L%7D%7Br_L%20tan%5Calpha%7D)
a)
![F_R = 238N\\ F_N = 1310N](https://tex.z-dn.net/?f=F_R%20%3D%20238N%5C%5C%20F_N%20%3D%201310N)
b)
![F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3](https://tex.z-dn.net/?f=F_R%20%3D%20%5Cmu%20F_N%5C%5C%20%5Cmu%20%3D%20%5Cfrac%7BF_R%7D%7BF_N%7D%20%5C%5C%20%5Cmu%20%3D%200.3)
c) Using the result from b and solving for ![r_F](https://tex.z-dn.net/?f=r_F)
![\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m](https://tex.z-dn.net/?f=%5C%5C%20%5Cmu%20%3D%200.15%5C%5C%20F_R%20%3D%20%5Cmu%20F_N%5C%5C%20r_F%20%3D%202.4m)