AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Consider the figure shown below.Write the force equation for both the blocks.
mg‌cos‌30°−T=ma
T=m(g‌cos‌30°−a) …… (I)
T−mg‌cos‌60°=ma …… (II)
From equation (I) and (II), we get
ma=m(g‌cos‌30°−a)−mg‌cos‌60°
a=

(√3−1)

4

g
The expression of the magnitude of the acceleration at centre of mass is,
acm=|

ma1+ma2

m+m

|
acm=|

a1+a2

2

|
Substitute values.
acm=|

(√3−1) 4 g ^ i + (√3−1) 4 g ^ j

2

|
acm=

g

2

√(

√3−1

4

)2+(

√3−1

4

)2
acm=(

√3−1

4√2

) g