Answer:
a). M = 20.392 kg
b). am = 0.56 [tex]m/s^2[/tex] (block), aM = 0.28 [tex]m/s^2[/tex] (bucket)
Explanation:
a). We got N = mg cos θ,
f = [tex]$\mu_s N$[/tex]
= [tex]$\mu_s mg \cos \theta$[/tex]
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + [tex]$\mu_s mg \cos \theta$[/tex] .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
[tex]$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$[/tex]
[tex]$M=2(m \sin \theta + \mu_s mg \cos \theta)$[/tex]
[tex]$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$[/tex]
M = 20.392 kg
b). [tex]$(h-x_m)+(h-x_M)+(h'+x_M)=l$[/tex] .............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
[tex]$-\ddot{x}-2\ddot x_M=0$[/tex]
[tex]$\ddot x_M=\frac{\ddot x_m}{2}$[/tex]
[tex]$a_M=\frac{a_m}{2}$[/tex] .....................(iv)
We got, N = mg cos θ
[tex]$f_K=\mu_K mg \cos \theta$[/tex]
∴ [tex]$T-(mg \sin \theta + f_K) = ma_m$[/tex]
[tex]$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$[/tex] ................(v)
Mg - 2T = M[tex]a_M[/tex]
[tex]$Mg-Ma_M=2T$[/tex]
[tex]$Mg-\frac{Ma_M}{2} = 2T$[/tex] (from equation (iv))
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}=T$[/tex] .....................(vi)
Putting (vi) in equation (v),
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$[/tex]
[tex]$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$[/tex]
[tex]$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$[/tex]
[tex]$a_m= 0.56 \ m/s^2$[/tex]
Using equation (iv), we get,
[tex]a_M= 0.28 \ m/s^2[/tex]
