Answer to Question #187452 in Mechanical Engineering for Manoj Kumar Duvvuru

Question

Q2: Find (i) acceleration of the masses and (ii) tension in the two strings. Take coefficient of friction for the contact surfaces of bodies A and B as 0.3. 15 kg B T A 12 kg A T B 20 T A C T B 45 kg

Answer 1

The constraint relation between blocks A,B and C gives,

$Ab = \frac{Aa}{2} + \frac{Ac}{2}$ (where A is acceleration of block)

Case 1

All blocks are in rest

$T - fa = 0$

$T - fc = 0$

$2T = Mb * g$

So $T = 10 * \frac{10}{2} = 50 N$

so $fa = fc = 50 N$

But $Max fa = 5 * 10 * 0.24 = 12 N$

so since required fa > max fa

all blocks cannot be in rest position

Case 2

Block C is in rest

T - fa = ma * Aa

T - fc = 0

mb * g - 2T = mb * Ab

$fa = ma * g * uk = 5 * 10 * 0.2 = 10 N$

T - 10 = 5 * Aa

10 * 10 - 2T = 10 * Ab

but Ab = Aa/2 + 0/2 = Aa/2

T - 10 = 5 * Aa ---(1)

100 - 2T = 5 * Aa ---(2)

2 * (1) + (2)

100 - 20 = 10 * Aa + 5 * Aa

15 * Aa = 80

Aa = 16/3 m/s

so

T = 5 * 16/3 + 10 = 110/3

so

fc = 110/3 = 36.67 N

but

Max fc = 10 * 10 * 0.24 = 24 N

So Block C cannot be in rest

CASE 3

ALL blocks are in motion

For Block A

T - fa = ma * Aa

T - 5 * 10 * 0.2 = 5 * Aa  

T - 10 = 5 * Aa ---(1)

For Block C

T - fc = mc * ac

T - 10 * 10 * 0.2 = 10 * Ac

T - 20 = 10 * Ac ----(2)

For Block B

mb g - 2T = mb * Ab

10 * 10 - 2T = 10 * Ab

100 - 2T = 10 * Ab

but Ab = Aa/2 + Ac/2

100 - 2T = 5 * Aa + 5 * Ac ----(3)

$3 * (2) + (3) - (1)$ gives

$100 + 10 - 3 * 20 = 3 * 10 * Ac + 5 * Ac$

35 * Ac = 50

Ac = 10/7 m/s

so

$T = 20 + 10 * \frac{10}{7} = 240/7 N$

$Aa = \frac{(240/7 - 10)}{5} = 34/7 m/s$

$Ac =\frac{ (\frac{10}{7} + \frac{34}{7})}{2} = \frac{22}{7} m/s$