2. 1st Case
Three Blocks of masses 2.0, 4.0,
and 6.0 Kg, arranged in the order
lower, middle and upper,
respectively, are connected by
strings on a frictionless inclined
plane 60°,A force of 120 N is
applied upward along the incline
to the uppermost block, causing
an upward movement of the
blocks.The connecting cords are
light. What is the acceleration of
the blocks? 60°
3. Problem Analysis
• M1 = 2,0 Kg
• M2 = 4,0 Kg
• M3 = 6,0 Kg
Applying Newton’s Second Law to each
Block We Have :
1. F – T2 – M3 g Sin 60° = M3 a
2. T2 – T1 – M2 g sin 60° = M2 a
3. T1 – M1 g sin 60° = M1 a
Then we adding those equation , so we will
get the value of acceleration in that
motion.
- Add the 3rd equation to 2nd then to the
1st in through tension equation.
60°
4. By subtitute those equation we will
get :
F-(M1+M2+M3) g sin60°= (M1+M2+M3)a
Insert the Value each of it:
120 – (12) 9.8 0.866 = 12 a
Acceleration = 1.51 m/s2
So, the acceleration of the blocks in
that motion is 1.51 m/s2
60°
