Learn acceleration of block by virtual work method & visual idea of constraint motion. Practice to find constraint acceleration between two blocks and an accelerating pulley.
A \(a\)
B \(2a\)
C \(\dfrac{a}{2}\)
D zero
\(\sum \;\vec{T.}\vec{v} = 0\)
Where \(\vec{T}\) is the tension in the string
\(\vec{v}\) is the velocity of block
T.v_{1} = T v_{1} cos 0° = T v_{1}
\(\vec{T.} \vec{v_2} = T \;v_2\; cos 180° = -Tv_2\)
\(\sum T v = 0\)
\(\Rightarrow Tv_1 – Tv_2 = 0\)
\(v_1 = v_2 \)
Similarly, \(\dfrac{dv_1}{dt}\) = \(\dfrac{dv_2}{dt}\)
\(\Rightarrow a_1 =a_2\)
A 6 m/sec2
B 2 m/sec2
C 8 m/sec2
D 3 m/sec2
Using Virtual Work Method
→ For pulley
= +2Ta
→ For block m_{1 }
= + Ta_{1}
→ For block m_{2 }
= +Ta_{2}
\(\sum\;Ta = 0\)
⇒ 2Ta + Ta_{1} + Ta_{2} = 0
⇒ 2a + a_{1} + a_{2} = 0
\(\sum \;\vec{T.}\vec{v} = 0\)
where \(\vec{T}\) is the tension in the string
\(\vec{v}\) is the velocity of block
T.v_{1} = Tv_{1} cos 0° = Tv_{1}
\(\vec{T.} \vec{v_2} = T \,v_2\; cos 180° = -Tv_2\)
\(\sum T v = 0\)
\(\Rightarrow Tv_1 – Tv_2 = 0\)
\(v_1 = v_2 \)
\(\Rightarrow a_1 =a_2\)
A 2a1 + a2 + a3 = 0
B 2a1 + a2 – a3 = 0
C 2a1 – a2 – a3 = 0
D 2a1 – a2 + a3 = 0
A 2a1 – a2 – a3 = 0
B a1 – a2 – a3 = 0
C a1 + a2 – a3 = 0
D 2a1 – 2a2 = a3
\(\sum \;\vec{T.}\vec{v} = 0\)
where \(\vec{T}\) is the tension in the string
\(\vec{v}\) is the velocity of block
T.v_{1} = Tv_{1} cos 0° = Tv_{1}
\(\vec{T.} \vec{v_2} = T \,v_2\; cos 180° = -Tv_2\)
\(\sum T v = 0\)
\(\Rightarrow Tv_1 – Tv_2 = 0\)
\(v_1 = v_2 \)
Similarly, \(\dfrac{dv_1}{dt}\) = \(\dfrac{dv_2}{dt}\)
\(\Rightarrow a_1 =a_2\)
A \(\dfrac{a}{2}\)
B \(\dfrac{a}{3}\)
C \(2{a}\)
D \(a\)