Laws of MotionHard
Question
In the system shown in figure mA = 4m, mB = 3m and mC = 8m. Friction is absent everywhere. String is light and inextensible. If the system is released from rest find the acceleration of block B
Options
A.
(leftward)
(leftward) B.
(leftward)
(leftward) C.
(rightward)
(rightward) D.
(rightward)
(rightward) Solution
a = b + c [string constrained]
T = mBb [Newtons II law for B in horizontal direction]
T = [mA + mc] c [Newton′s I law for A and C in horizontal direction]
mAg − T = mAa [Newton′s II law for A in vertical direction].
⇒ T =
⇒ b = 
Create a free account to view solution
View Solution FreeMore Laws of Motion Questions
When a mango tree is shaken mangoes are detached from the tree (branches) because :-...In the arrangement show in figure pulley A and B are mass less and the thread is inextensible. Mass of pulley C is equal...The ratio of gravitational mass to inertial mass is equal to :...The ratio of the inertial to gravitational mass is equal to :-...A block of mass 0.1 kg is held aganist a wall applying force of 5N on the block. If the coefficient of friction between ...