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

Question #187452

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


1
Expert's answer
2021-05-21T03:44:03-0400

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"




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