Question

2.5
Tension in rod at a distance 10 cm from end A is-
10cm 20cm
ΑΣ
B
32N
20N
(1) 18 N
(3) 24 N
(2) 20 N
(4) 36 N​

Answer 1

Answer:

hi nanba

my greetings to you

but I don't know how to do this sum

Answer 2

Answer:

according to Newton's law of motion,

body moves where force exists . here in right side force { 32N } is greater than left side side {20N}. so, body of mass 3kg moves in rightward.

Let the force exerted by 20cm part of the rod on the 10cm part is N.

first we have to find mass of 20cm part and 10cm part of body .

mass of 20cm part , M = 3kg × 20cm/(10cm + 20cm) = 2kg

mass of 10cm part , m = 3kg - 2kg = 1kg

now, use Newton's 2nd law ,

for 20cm part ,

forward force - backward force = Ma

32 - N = 2a -------(i) [ here a is acceleration]

similarly for 10cm part,

N - 20 = a -------(ii)

from eqs. (i) and (ii),

12 = 3a => a = 4 m/s²

now, N = 32 - 2a = 32 - 2 × 4 = 24

hence, required force = 24N