Question

The masses of the blocks A, B, and C shown in Fig. (a) are 4 kg, 2kg, and 2kg, respectively. Block A moves with an acceleration of 2.5 $m{s}^{-2}$.
a. Block C is removed from its position and placed on block A, shown in fig (b). What is now the acceleration of block C?
b. The positions of the blocks A and B is subsequently interchanged. Find the new acceleration of C. The coefficient of friction is the same for all the contact surface.

Answer 1

Given acceleration of the system, $a=2.5m{s}^{-2}$

For B and C : $2g+2g-T=(2+2)a$ (i)

For A: $T-\mu N=4a\Rightarrow T-\mu 4g=4a$ (ii)

Solving (i) and (ii), we get $\mu =0.5$

a. Let A and C move together with common accleration a. Then the acceleration of B is also a.

$2g-T=2a$ (i)

$T-\mu 6g=6a$ (ii)

Solving these equations, a comes out equations, a comes out to be negative, which is not possible. It means system will remain at rest and acceleration of C will be zero.

b. Let B and C move together.

$4g-T=4a,T-4\mu g=4a$

Solving them: $a=2.5m{s}^{-2}$

Let f be the friction between B and C. Then

For C:

$f=2a=2\times 2.5=5N$

It is because this friction acceleration to C is given by friction f. Limiting friction between B and C: $f_1=\mu 2g=\left(1/2\right)\times 2g=10N$. Since$f<f_t$, so C will not slip on B.

Hence, acceleration of $C=2.5m{s}^{-2}$