AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

The free body diagram of the two masses is shown below. The friction on mass ‘m’ is,
f1=µN1
f1=

2

3

mg‌cos‌45°
f1=

√2

3

mg
The friction on mass ′2m′ is,
f2=µN2
f2=

2

3

(2m)g‌cos‌45°
f2=

2√2

3

mg
The second law of motion for m mass,
T−f1−mg‌cos‌45°=ma
T−

√2

3

mg−

mg

√2

=ma ...... (I)
The second law of motion for 2m mass,
2mg‌cos‌45°−T−f2=2ma
√2mg−T−

2√2

3

mg=2ma ...... (II)
Solve equation (I) and (II),
3T−2√2mg=0
T=

2√2mg

3