Question

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.
Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?

Answer 1

Answer:

a). M = 20.392 kg

b). am = 0.56 $m/s^2$ (block),  aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got  N = mg cos θ,

                  f = $\mu_s N$

                    = $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$   .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$  .............(iii)

   Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$   .....................(iv)

We got,   N = mg cos  θ

                $f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

  $T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$  ................(v)

Mg - 2T = M$a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$    (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$   .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$